Chapter 3: Trigonometric Functions

Exercise 3.1

Q.1: Calculate the radian measurement of the given degree measurement:

(i).  \(25\;^{\circ}\)

(ii).  \(240\;^{\circ}\)

(iii).  \(-47\;^{\circ} \; 30′\)

(iv).  \(520\;^{\circ}\)

 

Sol:

(i).  \(25\;^{\circ}\)

As we know that, \(180\;^{\circ}\) = π radian

Therefore, \(25\;^{\circ}\) = \(\frac{\pi }{180} \times 25\) radian = \(\frac{5\pi }{36}\) radian

Hence, \(25\;^{\circ}\) = \(\frac{5\pi }{36}\) radian

 

(ii).  \(240\;^{\circ}\)

As we know that \(180\;^{\circ}\) = π radian

Therefore, \(240\;^{\circ}\) = \(\frac{\pi }{180} \times 240\) radian = \(\frac{4\pi }{3}\) radian

Hence, \(240\;^{\circ}\) = \(\frac{4\pi }{3}\) radian

 

(iii).  \(-47\;^{\circ} \; 30′\)

= \(-47\frac{1}{2}\)

= \(\frac{-95}{2}\) degree

As we know that, \(180\;^{\circ}\) = π radian

Therefore, \(\frac{-95}{2}\) degree = \(\frac{\pi }{180}\times \frac{-95}{2}\) radian = \(\frac{-19 }{36\; \times \;2}\pi\) radian = \(\frac{-19 }{72}\pi\)

Hence, \(-47\;^{\circ} \; 30′\) = \(\frac{-19 }{72}\pi\)

 

(iv).  \(520\;^{\circ}\)

As we know that, \(180\;^{\circ}\) = n radian

Therefore, \(520\;^{\circ}\) = \(\frac{\pi }{180} \times 520\) radian = \(\frac{26\pi }{9}\) radian

Hence, \(520\;^{\circ}\) = \(\frac{26\pi }{9}\) radian

 

 

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = \(\frac{22}{7}\)]

(i) \(\frac{11}{16}\)

 

(ii) -4

 

(iii) \(\frac{5\pi }{3}\)

 

(iv) \(\frac{7\pi }{6}\)

 

Sol:

(i).  \(\frac{11}{16}\):

As we know, that Π Radian = 180°

Therefore, \(\frac{11}{16}\) radian = \(\frac{180}{\pi }\times \frac{11}{16}\) degree

= \(\frac{45\times 11}{\pi \times 4 }\) degree

= \(\frac{45 \times 11 \times 7}{22 \times 4 }\) degree

= \(\frac{315}{8}\) degree

= \(39\;\frac{3}{8}\) degree

= \(39\;^{\circ} \;+ \; \frac{3\times 60}{8}\) minute  [\(1^{\circ}\) = 60’]

= \(39^{\circ} \;+ \; 22′ \;+ \;\frac{1}{2}\) minute

= \(39^{\circ} \;+ \; 22′ \;+ \;30”\)  [1’ = 60’’]

 

(ii).  -4:

As we know, that Π Radian = 180°

Therefore, -4 radian = \(\frac{180}{\pi } \times (-4)\) degree

= \(\frac{180 \times 7(-4)}{22}\) degree

= \(\frac{-2520}{11}\) degree

= \(-229\frac{1}{11}\) degree

= \(-229^{\circ} \; + \; \frac{1\times 60}{11}\) minutes     [\(1^{\circ}\) = 60’]

= \(-229^{\circ} \; +\; 5′ +\;\frac{5}{11}\)

= \(-229^{\circ} \; +\; 5′ +\;27”\) [1’ = 60’’]

 

(iii).  \(\frac{5\pi }{3}\)

As we know, that Π Radian = 180°

Therefore, \(\frac{5\pi}{3}\) radian = \(\frac{180}{\pi }\times \frac{5\pi}{3}\) degree

= \(300^{\circ}\)

 

(iv).  \(\frac{7\pi }{6}\)

As we know, that Π Radian = 180°

Therefore, \(\frac{7\pi}{6}\) radian = \(\frac{180}{\pi }\times \frac{7\pi }{6}\)

= \(210^{\circ}\)

 

 

Q.3: In a minute, wheel makes 360 revolutions. Through how many radians does it turn in 1 second?

Answer:

No. of revolutions made in a minute = 360 revolutions

Therefore, no. of revolutions made in a second = \(\frac{360}{60}\) = 6

In one revolution, the wheel rotates an angle of 2π radian.

Therefore, in 6 revolutions, it will turn an angle of 12π radian.

 

 

Q.4: Calculate the degree measurement of the angle subtended at the centre of a circle of radius 100 m by an arc of length 22 m.

[Use π = \(\frac{22}{7}\)]

Answer:

As we know,

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, \(\theta = \frac{l}{r}\)

Given:

r = 100 m and L = 22 m

We have,

\(\theta = \frac{22}{100}\) radian

= \(\frac{180}{\pi }\times \frac{22}{100}\) degree

= \(\frac{180 \;\times \;7 \;\times \;22}{22 \; \times 100 }\) degree

= \(\frac{126}{10 }\) degree

= \(12\frac{3}{5}\) degree

= \(12^{\circ} \; 36′\) [\(1^{\circ}\) = 60’]

Therefore, the req angle is \(12^{\circ} \; 36′\)

 

 

Q.5: In a circle of diameter 40 m, the length of the chord 20 m. Find the length of minor arc of chord.

Answer:

13

Diameter of circle = 40 m

Radius of circle = \(\frac{40}{20}\) m = 20 m

Let XY be the chord (length = 20 m) of the circle.

In \(\Delta OXY\), OX = OY  = radius of the circle = 20 m

XY = 20 m

Therefore,

\(\Delta OXY\) is an equilateral triangle.

\(\theta \;= \;60^{\circ}\) = \(\frac{\pi }{3}\) radian

As we know,

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, θ = \(\frac{l}{r}\)

\(\frac{\pi }{3}\) = \(\frac{arc\;(AB\;)}{20}\)

\(\frac{arc\;(AB\;)}{20}\) = \(\frac{20\pi }{3}\) m

The length of the minor arc of the chord is \(\frac{20\pi }{3}\) m.

 

 

Q.6: In two circles, arcs which has same length subtended at an angle of \(60^{\circ}\) and \(75^{\circ}\) at the center. Calculate the ratio of their radii.

 Sol:

Let, the radii of two circles be \(r_{1}\) and \(r_{2}\).

Let, an arc of length l subtend an angle of \(60^{\circ}\) at the center of the circle of radius \(r_{1}\), while let an arc of length l subtend an angle of \(75^{\circ}\) at the center of the circle of radius \(r_{2}\).

\(\\60^{\circ}\) = \(\frac{\pi }{3}\) radian

\(75^{\circ}\) = \(\frac{5\pi }{12}\) radian

In a circle of radius r unit, if an arc of length l unit subtends an angle θ then:

θ = \(\frac{l}{r}\)

l = r θ

l = \(\frac{r_{1}\;\pi }{3}\)

l = \(\frac{r_{2}\;5\pi }{12}\)

\(\frac{r_{1}\;\pi }{3}\) = \(\frac{r_{2}\;5\pi }{12}\)

\(r_{1}\) = \(\frac{r_{2}5}{4}\)

\(\frac{r_{1}}{r_{2}}\) = \(\frac{5}{4}\)

Therefore, the ratio of radii is 5: 4

 

Q.7: Calculate the angle in radian through which a pendulum swings if the length is 75 cm and the tip describes an arc of length

(i) 10 cm

 

(ii) 15 cm

 

(iii) 21 cm

 

Sol:

As we know that, in a circle of radius ‘r’ unit, if an arc of length ‘l’ unit subtends an angle \(\theta\) radian at the center, then:

\(\theta = \frac{l}{r}\\\)

Here, r = 75 cm

(i).  10 cm:

\(\theta\) = \(\frac{10}{75}\) radian = \(\frac{2}{15}\) radian

 

(ii).  15 cm:

\(\theta\) = \(\frac{15}{75}\) radian = \(\frac{1}{5}\) radian

 

(iii).  21 cm:

\(\theta\) = \(\frac{21}{75}\) radian = \(\frac{7}{25}\) radian

 

 

Exercise 3.2

Q.1: Calculate the values of five trigonometric func. if \(\cos y\) = \(-\frac{1}{2}\) and y lies in 3rd quadrant.

Sol:

(i)  sec y :

Since, cos y = \(\frac{1}{2}\)

Therefore, sec y = \(\frac{1}{\cos y}\) = \(\frac{1}{\left (-\frac{1}{2} \right )}\)

Hence, sec y = -2

 

(ii)  sin y :

Since, \(\sin ^{2}y \;+ \;\cos ^{2}y \;= \;1\)

Therefore, \(\sin ^{2}y \;= \;1 \;- \;\cos ^{2}y\)

\(\Rightarrow\) \(\sin ^{2}y \;= \;1 \;- \;\left (-\frac{1}{2} \right )^{2}\)

\(\Rightarrow\) \(\sin ^{2}y \;= \;1 \;- \;\frac{1}{4}\)

\(\Rightarrow\) \(\sin ^{2}y \;= \;\frac{3}{4}\)

\(\Rightarrow\) \(\sin y \;= \;\pm \frac{\sqrt{3}}{2}\)

Since, y lies in the third quadrant, the value of sin y will be negative.

Therefore, sin y = \(\frac{\sqrt{3}}{2}\)

 

(iii)  cosec y = \(\frac{1}{\sin y}\) = \(\frac{1}{\left (-\frac{\sqrt{3}}{2} \right )}\) = \(-\frac{2}{\sqrt{3}}\)

Therefore, cosec y = \(-\frac{2}{\sqrt{3}}\)

 

(iv)  tan y = \(\frac{\sin y}{\cos y}\) = \(\tan y=\frac{\left (-\frac{\sqrt{3}}{2} \right )}{\left (-\frac{1}{2} \right )}\) = \(\sqrt{3}\)

Therefore, tan y = \(\sqrt{3}\)

 

(v)  cot y = \(\frac{1}{\tan y}\) = \(\frac{1}{\sqrt{3}}\)

Therefore, cot y = \(\frac{1}{\sqrt{3}}\)

 

 

Q.2: Calculate the other five trigonometric function if we are given the values for sin y = \(\frac{3}{5}\), where y lies in second quadrant.

Sol:

sin y = \(\frac{3}{5}\)

Therefore, cosec y = \(\frac{1}{Sin\; y}\) = \(\frac{1}{\frac{3}{5}} = \frac{5}{3}\)

Since, \(sin^{2}\;y \; + \; cos^{2}\;y = 1\)

\(\Rightarrow\)  \( cos^{2} y = 1 – sin^{2} y\)

\(\Rightarrow\)  \(cos^{2} y = 1 – \left ( \frac{3}{5} \right )^{2}\)

\(\Rightarrow\) \(cos^{2} y = 1 – \frac{9}{25}\)

\(\Rightarrow\) \(cos^{2} y = \frac{16}{25}\)

\(\Rightarrow\) cos y = \(\pm \frac{4}{5}\)

Since, y lies in the 2nd quadrant, the value of cos y will be negative,

Therefore, cos y = – \(\frac{4}{5}\)

\(\Rightarrow\) sec y = \(\frac{1}{cos \; y} = \frac{1}{\left (- \frac{4}{5} \right )} =\; – \frac{5}{4}\)

\(\Rightarrow\) tan y = \(\frac{sin \; y}{cos \; y} = \frac{\left (\frac{3}{5} \right ) }{\left (-\frac{4}{5} \right ) } = \;- \frac{3}{4}\)

\(\Rightarrow\) cot y = \(\frac{1}{tan \; y} =\;- \frac{4}{3}\)

 

 

 Q.3: Find the values of other five trigonometric functions if \(cot \; y = \frac{3}{4}\), where y lies in the third quadrant.

Sol:

cot y = \(\frac{3}{4}\)

Since, tan y = \(\frac{1}{cot \; y}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)

Since, \(1 + tan^{2}y = sec^{2}y\)

\(\Rightarrow\)  \(1 + \left ( \frac{4}{3} \right )^{2} = sec^{2}y\)

\(\Rightarrow\)  \(1 + \frac{16}{9} = sec^{2}y\)

\(\Rightarrow\)  \(\frac{25}{9} = sec^{2}y\)

\(\Rightarrow\)  sec y = \(\pm \frac{5}{3}\)

Since, y lies in the 3rd quadrant, the value of sec y will be negative.

Therefore, sec y = \(\;- \frac{5}{3}\)

cos y = \(\frac{1}{sec \; y} = \frac{1}{\left (- \frac{5}{3} \right )} = – \frac{3}{5}\)

Since, tan y = \(\frac{sin \; y}{cos \; y}\)

\(\Rightarrow\)  \( \frac{4}{3} = \frac{sin \; y}{\left (- \frac{3}{5} \right )}\)

\(\Rightarrow \) sin y = \(\left (\frac{4}{3} \right ) \times \left (- \frac{3}{5} \right ) = -\frac{4}{5}\)

\(\Rightarrow\)  cosec y = \(\frac{1}{sin \; y} = -\frac{5}{4}\)

 

 

Q.4: Find the values of other five trigonometric if \(sec \; y = \frac{13}{5}\), where y lies in the fourth quadrant.

Sol:

sec y = \(\frac{13}{5}\)

cos y = \(\frac{1}{sec \; y} = \frac{1}{\left (\frac{13}{5}\right )} = \frac{5}{13}\)

Since, \(sin^{2}y + cos^{2}y = 1\)

\(\Rightarrow\)   \(sin^{2}y = 1 – cos^{2}y\)

\(\Rightarrow\)   \(sin^{2}y = 1 – \left ( \frac{5}{13} \right )^{2}\)

\(\Rightarrow\)   \(sin^{2}y = 1 – \frac{25}{169} = \frac{144}{169}\)

\(\Rightarrow\) sin y = \(\pm \frac{12}{13}\)

Since, y lies in the 4th quadrant, the value of sin y will be negative.

Therefore, sin y = \(– \frac{12}{13}\)

\(\Rightarrow\) cosec y = \(\frac{1}{sin \; y} = \frac{1}{\left (- \frac{12}{13} \right )} = \;- \frac{13}{12}\)

\(\Rightarrow\) tan y = \(\frac{sin \; y}{cos \; y} = \frac{\left (- \frac{12}{13} \right )}{\left (\frac{5}{13} \right )} =\; -\frac{12}{5}\)

\(\Rightarrow\) cot y = \(\frac{1}{tan \; y} = \frac{1}{\left ( -\frac{12}{5} \right )} = \;-\frac{5}{12}\)

 

 

Q.5: Find the values of other five trigonometric function if tan y = \(– \frac{5}{12}\) and y lies in second quadrant.

Sol:

tan  y = \(-\frac{5}{12}\)   [Given]

And, cot y = \(\frac{1}{tan \; y} = \frac{1}{ \left (- \frac{5}{12} \right )} = – \frac{12}{5}\)

Since, \(1 + tan^{2} y = sec^{2} y\)

Therefore, \(1 + \left ( – \frac{5}{12} \right )^{2} = sec^{2} y\)

\(\Rightarrow\)   \(sec^{2} y = 1 + \left ( \frac{25}{144} \right )\)

\(\Rightarrow \)  \(sec^{2} y = \frac{169}{144}\)

\(\Rightarrow\) sec y = \(\pm \frac{13}{12}\)

Since, y lies in the 2nd quadrant, the value of sec y will be negative.

Therefore, sec  y = \(– \frac{13}{12}\)

cos y = \(\frac{1}{sec \; y } = \frac{1}{\left (- \frac{13}{12} \right )} = \left (- \frac{12}{13} \right )\\\)

Since, \(tan \; y = \frac{sin \; y}{cos \; y}\)

\(\Rightarrow\)  \( -\frac{5}{12} = \frac{sin \; y}{\left (- \frac{12}{13} \right )}\\\)

\(\Rightarrow\) sin y = \(\left (-\frac{5}{12} \right ) \times \left (- \frac{12}{13}\right ) = \frac{5}{13}\\\)

\(\Rightarrow\) cosec y = \(\frac{1}{sin \; y} = \frac{1}{\left ( \frac{5}{13} \right )} = \frac{13}{5}\)

 

 

Q.6: Calculate the value of trigonometric function sin 765°.

Sol:

The values of sin y repeat after an interval of 360° or 2n.

Therefore, sin 765° = sin ( 2 × 360° + 45° ) = sin 45° = \(\frac{1}{\sqrt{2}}\)

 

 

Q.7: Calculate the value of trigonometric function cosec [-1410°]

 Sol:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, cosec [-1410°] = cosec [-1410° + 4 × 360°] = cosec [ -1410° + 1440°] = cosec 30° = 2

 

 

Q.8: Calculate the value of the trigonometric function \(\tan \frac{19\pi }{3}\).

Sol:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, \(\tan \frac{19\pi }{3}\) = \(\tan 6\frac{1}{3}\pi\) = \(\tan \left ( 6\pi + \frac{\pi }{3} \right )\) = \(\tan \frac{\pi }{3}\) = tan 60° = \(\sqrt{3}\)

 

 

Q.9: Calculate the value of the trigonometric function \(\sin \left ( -\frac{11\pi }{3} \right )\).

 Answer:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, \(\sin \left ( -\frac{11\pi }{3} \right )\) = \(\sin \left ( -\frac{11\pi }{3} + 2 \times 2\pi \right )\) = \(\sin \frac{\pi }{3}\) = \(\frac{\sqrt{3}}{2}\)

 

 

Q.10: Calculate the value of the trigonometric function \(\cot \left ( -\frac{15\pi }{4} \right )\).

 Sol:

It is known that the values of \(\tan y\) repeat after an interval of \(180^{\circ}\) or n.

Therefore, \(\cot \left ( -\frac{15\pi }{4} \right )\) = \(\cot \left ( -\frac{15\pi }{4} + 4\pi \right )\) = \(\cot \frac{\pi }{4}\) = 1

 

 

Exercise 3.3

Q.1: Prove:

\(\sin ^{2} \frac{\pi }{6} + \cos ^{2} \frac{\pi }{3} – \tan ^{2} \frac{\pi }{4} = -\frac{1}{2}\)

 

Sol:

Now, taking L.H.S.

\(\sin ^{2} \frac{\pi }{6} + \cos ^{2} \frac{\pi }{3} – \tan ^{2} \frac{\pi }{4}\):

= \(\\\left (\frac{1}{2} \right )^{2} + \left (\frac{1}{2} \right )^{2} – \left ( 1 \right )^{2}\)

= \(\\\frac{1}{4} + \frac{1}{4} – 1\) = \(-\frac{1}{2}\)

= R.H.S.

 

 

Q.2: Prove:

\(2 \sin ^{2} \frac{\pi }{6} + cosec^{2}\frac{7\pi }{6} \cos ^{2}\frac{\pi }{3} = \frac{3}{2}\)

 

Sol:

Now, taking L.H.S.

\(2 \sin ^{2} \frac{\pi }{6} + cosec^{2}\frac{7\pi }{6} \cos ^{2}\frac{\pi }{3}\\\) :

= \(\\2 \left (\frac{1}{2} \right )^{2} + cosec ^{2}\left ( \pi + \frac{\pi }{6} \right )\left ( \frac{1}{2} \right )^{2}\)

= \(2 \times \frac{1}{4} + \left (-cosec \frac{\pi }{6} \right )^{2}\left ( \frac{1}{4} \right )\)

= \(\frac{1}{2} + \left ( -2 \right )^{2}\left ( \frac{1}{4} \right )\)

= \(\frac{1}{2} + \frac{4}{4}\)

= \(\frac{1}{2} + 1\)

= \(\frac{3}{2}\)

= R.H.S.

 

 

Q.3: Prove:

\(\cot ^{2}\frac{\pi }{6} + cosec \frac{5\pi }{6} + 3\tan ^{2} \frac{\pi }{6} = 6\)

 

Sol:

Taking L.H.S.

\(\cot ^{2}\frac{\pi }{6} + cosec \frac{5\pi }{6} + 3\tan ^{2} \frac{\pi }{6}\\\) :

= \(\left (\sqrt{3} \right )^{2} + cosec \left ( \pi – \frac{\pi }{6} \right ) + 3\left ( \frac{1}{\sqrt{3}} \right )^{2}\)

= \(3 + cosec \frac{\pi }{6} + 3\times \frac{1}{3}\)

= 3 + 2 + 1 = 6

= R.H.S.

 

 

Q.4: Prove:

\(2\sin ^{2} \frac{3\pi }{4} + 2\cos ^{2} \frac{\pi }{4} + 2\sec ^{2} \frac{\pi }{3} = 10\)

 

Sol:

Now, taking L.H.S.

\(2\sin ^{2} \frac{3\pi }{4} + 2\cos ^{2} \frac{\pi }{4} + 2\sec ^{2} \frac{\pi }{3}\\\) :

= \(\\2\left \{ \sin \left ( \pi – \frac{\pi }{4} \right ) \right \}^{2} + 2\left (\frac{1}{\sqrt{2}} \right )^{2} + 2\left ( 2 \right )^{2}\)

= \(2\left \{ \sin \frac{\pi }{4} \right \}^{2} + 2\times \frac{1}{2} + 8\)

= \(2\left ( \frac{1}{\sqrt{2}} \right )^{2}\) + 1 + 8

= 1 + 1 + 8 = 10

= R.H.S.

 

 

Q.5: Calculate the value of:

(i).  \(\sin 75^{\circ}\)

 

(ii).  \(\tan 15^{\circ}\)

 

Sol:

(i).  \(\sin 75^{\circ}\):

= \(\sin \left ( 45^{\circ} + 30^{\circ} \right )\)

= \(\sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ}\)

Since, [sin (x + y) = sin x cos y + cos x sin y]

= \(\left (\frac{1}{\sqrt{2}} \right )\left (\frac{\sqrt{3}}{2} \right ) + \left (\frac{1}{\sqrt{2}} \right )\left ( \frac{1}{2} \right )\)

= \(\frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}\)

= \(\frac{\sqrt{3} + 1}{2\sqrt{2}}\)

 

(ii).  \(\tan 15^{\circ}\):

= \(\tan \left ( 45^{\circ} – 30^{\circ}\right )\)

= \(\frac{\tan 45^{\circ} – \tan 30^{\circ}}{1 + \tan 45^{\circ}\tan 30^{\circ}}\)

Since, [tan (x – y) = \(\frac{\tan x – \tan y}{1 + \tan x \tan y}\)]

= \(\\\frac{1 – \frac{1}{\sqrt{3}}}{1 + 1\left ( \frac{1}{\sqrt{3}} \right )}\)

= \(\\\frac{\frac{\sqrt{3} – 1}{\sqrt{3}}}{\frac{\sqrt{3 + 1}}{\sqrt{3}}}\)

= \(\\\frac{\sqrt{3} – 1}{\sqrt{3} + 1}\)

= \(\\\frac{\left ( \sqrt{3} – 1\right )^{2}}{\left ( \sqrt{3} + 1\right )\left ( \sqrt{3} – 1 \right )}\)

= \(\\\frac{3 + 1 – 2\sqrt{3}}{\left ( \sqrt{3} \right )^{2} – \left ( 1 \right )^{2}}\)

= \(\frac{4 – 2\sqrt{3}}{3 – 1}\)

= \(2 – \sqrt{3}\)

 

 

Q.6:Prove:

\(\cos \left ( \frac{\pi }{4} – x\right ) \cos \left ( \frac{\pi }{4} – y\right ) – \sin \left ( \frac{\pi }{4} – x\right ) \sin \left ( \frac{\pi }{4} – y\right ) = \sin \left ( x+y \right )\)

 

Sol:

Now, taking L.H.S.

\(\cos \left ( \frac{\pi }{4} – x\right ) \cos \left ( \frac{\pi }{4} – y\right ) – \sin \left ( \frac{\pi }{4} – x\right ) \sin \left ( \frac{\pi }{4} – y\right ):\\\)

= \(\\\frac{1}{2}\left [ 2\cos \left ( \frac{\pi }{4} – x \right ) \cos \left ( \frac{\pi }{4} – y \right ) \right ] + \frac{1}{2}\left [ -2\sin \left ( \frac{\pi }{4} – x \right ) \sin \left ( \frac{\pi }{4} – y \right ) \right ]\\\)

\(=\frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi }{4} – x \right ) + \left ( \frac{\pi }{4} – y \right )\right \} + \cos \left \{ \left ( \frac{\pi }{4} – x \right ) – \left ( \frac{\pi }{4} – y \right )\right \}\right ] + \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi }{4} – x \right ) + \left ( \frac{\pi }{4} – y \right )\right \} – \cos \left \{ \left ( \frac{\pi }{4} – x \right ) – \left ( \frac{\pi }{4} – y \right )\right \}\right ]\\\)

Since, [2cos A cos B = cos (A + B) + cos (A – B)]

And, [2sin A sin B = cos (A + B) – cos (A – B)]

= \(\\2\times \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi }{4} – x \right ) + \left ( \frac{\pi }{4} – y \right ) \right \} \right ]\\\)

= \(\\\cos \left [ \frac{\pi }{2} – \left ( x + y \right )\right ]\)

= sin (x + y)

= R.H.S.

 

 

Q.7: Prove:

\(\frac{\tan \left ( \frac{\pi }{4} + x\right )}{\tan \left ( \frac{\pi }{4} – x\right )} = \left (\frac{1 + \tan x}{1 – \tan x} \right )^{2}\)

 

Sol:

Since, tan (A + B)= \(\frac{\tan A + \tan B}{1 – \tan A\tan B}\\\)

And, tan (A – B) = \(\frac{\tan A – \tan B}{1 + \tan A\tan B}\)

Now, taking L.H.S.

\(\frac{\tan \left ( \frac{\pi }{4} + x\right )}{\tan \left ( \frac{\pi }{4} – x\right )}\\\):

= \(\\\frac{\left (\frac{\tan \frac{\pi }{4} + \tan x}{1 – \tan \frac{\pi }{4} \tan x} \right )}{\left (\frac{\tan \frac{\pi }{4} – \tan x}{1 + \tan \frac{\pi }{4} \tan x} \right )}\\\)

= \(\\\frac{\left (\frac{1 + \tan x}{1 – \tan x} \right )}{\left (\frac{1 – \tan x}{1 + \tan x} \right )}\\\)

= \(\\\left (\frac{1 + \tan x}{1 – \tan x} \right )^{2}\)

= R.H.S.

 

 

Q.8: Prove:

\(\frac{\cos \left ( \pi + x \right ) \cos \left ( -x \right )}{\sin \left ( \pi – x \right ) \cos \left ( \frac{\pi }{2} + x \right )} = \cot ^{2} x\)

 

Sol:

Now, taking L.H.S.

\(\frac{\cos \left ( \pi + x \right ) \cos \left ( -x \right )}{\sin \left ( \pi – x \right ) \cos \left ( \frac{\pi }{2} + x \right )}\\\):

= \(\\\frac{\left [ -\cos x \right ]\left [ \cos x \right ]}{\left ( \sin x \right )\left ( -\sin x \right )}\\\)

= \(\\\frac{-\cos ^{2} x}{-\sin ^{2} x}\) = \(\cot ^{2} x\)

= R.H.S.

 

 

Q.9: Prove:

\(\cos \left ( \frac{3\pi }{2} + x\right ) \cos \left ( 2\pi + x \right )\left [ \cot \left ( \frac{3\pi }{2} – x\right ) + \cot \left ( 2\pi + x \right ) \right ] = 1\)

 

Sol:

Now, taking L.H.S.

\(\cos \left ( \frac{3\pi }{2} + x\right ) \cos \left ( 2\pi + x \right )\left [ \cot \left ( \frac{3\pi }{2} – x\right ) + \cot \left ( 2\pi + x \right ) \right ]\\\):

= \(\\\sin x\cos x \left [ \tan x + \cot x \right ]\\\)

= \(\\\sin x\cos x \left ( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right )\\\)

= \(\\\left (\sin x\cos x \right )\left [ \frac{\sin ^{2} x + \cos ^{2} x}{\sin x \cos x} \right ]\\\)

= \(\\\sin ^{2} x + \cos ^{2} x\) = 1

= R.H.S.

 

 

Q.10: Prove:

\(\sin \!\left ( n + 1 \right )\!x \; \sin\! \left ( n + 2 \right )\!x + \cos \! \left ( n + 1 \right )\! x \cos \! \left ( n + 2 \right )\! x = \cos x\)

 

Sol:

Now, taking L.H.S.

\(\sin \!\left ( n + 1 \right )\!x \; \sin\! \left ( n + 2 \right )\!x + \cos \! \left ( n + 1 \right )\! x \cos \! \left ( n + 2 \right )\! x:\\\)

=\(\\\frac{1}{2}\left [ 2\sin \! \left ( n + 1 \right ) \! x \sin \! \left ( n + 2 \right ) \! x + 2\cos \! \left ( n + 1 \right ) \! x \cos \! \left ( n + 2 \right ) \! x\right ]\\\)

=\(\\\frac{1}{2}\left [ \cos \left \{ \left ( n + 1 \right )x – \left ( n + 2 \right )x \right \} – \cos \left \{ \left ( n + 1 \right )x + \left ( n + 2 \right )x \right \} + \cos \left \{ \left ( n + 1 \right )x + \left ( n + 2 \right )x \right \} + \cos \left \{ \left ( n + 1 \right )x – \left ( n + 2 \right )x \right \}\right ]\\\)

Since, [2sin A sin B = cos (A + B) – cos (A – B)]

And, [2cos A cos B = cos (A + B) + cos (A – B)]

= \(\\\frac{1}{2}\times 2 \cos \left \{ \left ( n + 1 \right )x – \left ( n + 2 \right )x \right \}\\\)

= \(\\\cos \left ( -x \right )\) = cos x

= R.H.S.

 

 

Q.11 Prove:

\(\cos \left ( \frac{3\pi }{4} + x\right ) – \cos \left ( \frac{3\pi }{4} – x\right ) = -\sqrt{2}\sin x\)

 

Sol:

Since, cos A – cos B = \(-2\sin \left ( \frac{A + B}{2} \right )\sin \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\cos \left ( \frac{3\pi }{4} + x\right ) – \cos \left ( \frac{3\pi }{4} – x\right ):\\\)

= \(\\-2\sin \left \{ \frac{\left ( \frac{3\pi }{4} + x \right ) + \left ( \frac{3\pi }{4} – x \right )}{2} \right \} sin \left \{ \frac{\left ( \frac{3\pi }{4} + x \right ) – \left ( \frac{3\pi }{4} – x \right )}{2} \right \}\\\)

= \(\\-2\sin \left ( \frac{3\pi }{4} \right )\sin x\)

= \(\\-2\sin \left ( \pi – \frac{\pi }{4} \right )\sin x\)

= \(\\-2\sin \frac{\pi }{4} \sin x\)

= \(\\-2 \times \frac{1}{\sqrt{2}} \times \sin x\)

= \(\\-\sqrt{2} \sin x\)

= R.H.S.

 

 

Q.12: Prove:

\(\sin ^{2} 6x – \sin ^{2} 4x = \sin\! 2x \; \sin \! 10x\)

 

Sol:

Since, sin A + sin B = \(2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

And, sin A – sin B = \(2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\sin ^{2} 6x – \sin ^{2} 4x\\\):

= \(\\\left (\sin 6x + \sin 4x \right )\left ( \sin 6x – \sin 4x \right )\)

= \(\\\left [ 2\sin \left ( \frac{6x + 4x}{2} \right ) \cos \left ( \frac{6x – 4x}{2} \right ) \right ] \left [ 2\cos \left ( \frac{6x + 4x}{2} \right ) \sin \left ( \frac{6x – 4x}{2} \right ) \right ]\)

= (2sin 5x cos x) (2cos 5x sin x)

= (2 sin 5x cos 5x) (2 cos x sin x)

= sin 10x sin 2x

= R.H.S.

 

 

Q.13: Prove:

\(\cos ^{2} 2x – \cos ^{2} 6x = \sin 4x \sin 8x\)

 

Sol:

Since, cos A + cos B = \(2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

And, cos A – cos B = \(2\sin \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\cos ^{2} 2x – \cos ^{2} 6x\):

= (cos 2x + cos 6x)  (cos 2x – cos 6x)

= \(\\\left [ 2\cos \left ( \frac{2x + 6x}{2} \right ) \cos \left ( \frac{2x – 6x}{2} \right ) \right ] \left [ -2 \sin \left ( \frac{2x + 6x}{2} \right ) \sin \left ( \frac{2x – 6x}{2} \right ) \right ]\\\)

= [2cos 4x cos (-2x)]   [ -2sin 4x sin ( -2x)]

= [2cos 4x cos 2x]  [-2sin 4x ( -sin2x)]

= [2sin 4x cos 4x]  [2sin 2x cos 2x]

= sin 8x sin 4x

= R.H.S.

 

 

Q.14:Prove:

\(\sin 2x + 2\sin 4x + \sin 6x = 4 \cos ^{2} x \sin 4x\)

 

Sol:

Now, taking L.H.S.

sin 2x + 2sin 4x + sin 6x:

= [sin 2x + sin 6x] + 2 sin 4x

= \(\\\left [ 2 \sin \left (\frac{2x + 6x}{2} \right ) \left (\frac{2x – 6x}{2} \right ) \right ] + 2 \sin 4x\\\)

Since, sin A + sin B = \(2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

= 2sin 4x cos(-2x) + 2sin 4x

= 2sin 4x cos 2x + 2sin 4x

= 2sin 4x (cos 2x + 1)

= \(2 \sin 4x \; \left ( 2 \cos ^{2} x – 1 + 1\right )\)

= \(2 \sin 4x \; \left ( 2 \cos ^{2} x \right )\)

= \(4 \cos ^{2} x \sin 4x\)

= R.H.S.

 

 

Q.15: Prove:

\(\cot 4x \left ( \sin 5x + \sin 3x \right ) = \cot x \left ( \sin 5x – \sin 3x \right )\)

 

Sol:

Now, taking L.H.S.

cot 4x (sin 5x + sin 3x):

= \(\frac{\cos 4x}{\sin 4x} \left [ 2 \sin \left ( \frac{5x + 3x}{2} \right ) \cos \left ( \frac{5x – 3x}{2} \right ) \right ]\)

Since, sin A – sin B = \(2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

= \(\left ( \frac{\cos 4x}{\sin 4x} \right )\left [ 2 \sin 4x \; \cos x \right ]\)

=2cos 4x cos x . . . . . . . . . . . . . . . (1)

Now, taking R.H.S.

cot x (sin 5x – sin 3x):

= \(\frac{\cos x}{\sin x} \left [ 2 \cos \left ( \frac{5x + 3x}{2} \right ) \sin \left ( \frac{5x – 3x}{2} \right ) \right ]\)

\(\sin A – \sin B = 2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

= \(\frac{\cos x}{\sin x} \left [ 2 \cos 4x\; \sin x \right ]\)

= 2 cos 4x cos x . . . . . . . . . . . . . . . . . . . . (2)

From equation (1) and (2):

L.H.S. = R.H.S.

 

 

Q.16: Prove:

\(\frac{\cos 9x – \cos 5x}{\sin 17x – \sin 3x} = -\frac{\sin 2x}{\cos 10x}\)

 

Sol:

Since, cos A – cos B = \(2\sin \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

And, sin A – sin B = \(2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\frac{\cos 9x – \cos 5x}{\sin 17x – \sin 3x}:\\\)

= \(\\\frac{-2 \sin \left (\frac{9x + 5x}{2} \right )\; \sin \left (\frac{9x – 5x}{2} \right )}{2\cos \left (\frac{17x + 3x}{2} \right ) \; \sin \left (\frac{17x – 3x}{2} \right )}\\\)

= \(\\\frac{-2 \sin 7x \; \sin 2x}{2 \cos 10x \; \sin 7x}\\\)

= \(\\-\frac{\sin 2x}{\cos 10x}\)

= R.H.S.

 

 

Q.17: Prove:

\(\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x\)

 

Sol:

Since, sin A + sin B = \(2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

And, cos A + cos B = \(2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x}\\\):

= \(\\\frac{2 \sin \left ( \frac{5x + 3x}{2} \right ) \; \cos \left ( \frac{5x – 3x}{2} \right ) }{2 \cos \left ( \frac{5x + 3x}{2} \right ) \; \cos \left ( \frac{5x – 3x}{2} \right )}\\\)

= \(\\\frac{2 \sin 4x \cos x}{2 \cos 4x \cos x}\) = tan 4x

=R.H.S.

 

 

Q.18: Prove:

\(\frac{\sin x – \sin y}{\cos x + \cos y} = \tan \frac{x – y}{2}\)

 

Sol:

Since, cos A + cos B = \(2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

And, sin A – sin B = \(2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

Now taking L.H.S.

\(\frac{\sin x – \sin y}{\cos x + \cos y}:\\\)

= \(\\\frac{2 \cos \left ( \frac{x + y}{2} \right ) \;\sin \left ( \frac{x – y}{2} \right )}{2 \cos \left ( \frac{x + y}{2} \right ) \;\cos \left ( \frac{x – y}{2} \right )}\\\)

= \(\\\frac{\sin \left ( \frac{x – y}{2} \right )}{\cos \left ( \frac{x – y}{2} \right )}\)

= \(\\\tan \left ( \frac{x – y}{2} \right )\)

= R.H.S.

 

 

Q.19: Prove:

\(\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x\)

 

Sol::

Since, sin A + sin B = \(2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

And, cos A + cos B = \(2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\frac{\sin x + \sin 3x}{\cos x + \cos 3x}:\\\)

= \(\\\frac{2 \sin \left ( \frac{x + 3x}{2} \right ) \;\cos \left ( \frac{x – 3x}{2} \right )}{2 \cos \left ( \frac{x + 3x}{2} \right ) \;\cos \left ( \frac{x – 3x}{2} \right )}\)

= \(\\\frac{\sin 2x}{\cos 2x}\) = tan 2x

= R.H.S.

 

 

Q.20: Prove:

\(\frac{\sin x – \sin 3x}{\sin ^{2} x – \cos ^{2} x} = 2\sin x\)

 

Answer:

Since, sin A – sin B = \(2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

\(\cos ^{2} A – \sin ^{2} A = \cos 2A\\\)

Now, taking L.H.S.

\(\frac{\sin x – \sin 3x}{\sin ^{2} x – \cos ^{2} x}\\\):

= \(\\\frac{2 \cos \left (\frac{x + 3x}{2} \right ) \sin \left ( \frac{x – 3x}{2} \right )}{ – \cos 2x}\\\)

= \(\\\frac{2 \cos 2x \;\sin \left ( -x \right )}{-\cos 2x}\\\)

= \(\\-2x \left ( -\sin x \right )\) = 2 sin x

= R.H.S.

 

 

Q.21: Prove:

\(\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x\)

 

Sol:

Taking L.H.S.

\(\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}\):

= \(\\\frac{\left (\cos 4x + \cos 2x \right )+ \cos 3x}{\left (\sin 4x + \sin 2x \right ) + \sin 3x}\\\)

= \(\\\frac{2 \cos \left ( \frac{4x + 2x}{2} \right ) \cos \left ( \frac{4x – 2x}{2} \right ) + \cos 3x}{2 \sin \left ( \frac{4x + 2x}{2} \right ) \cos \left ( \frac{4x – 2x}{2} \right ) + \sin 3x}\\\)

\(\\\begin{bmatrix} \sin A + \sin B = 2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\\* \cos A + \cos B = 2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\\ \end{bmatrix}\)

= \(\frac{2 \cos 3x \; \cos x + \cos 3x}{2 \sin 3x \; \cos x + \sin 3x}\\\)

= \(\\\frac{\cos 3x \left ( 2 \cos x + 1 \right )}{\sin 3x \left ( 2 \cos x + 1 \right )}\) = cot 3x

= R.H.S.

 

 

Q.22: Prove:

\(\cot x \;\cot 2x – \cot 2x \;\cot 3x – \cot 3x \;\cot x = 1\)

 

Sol:

Now, taking L.H.S.

cot x cot 2x – cot 2x cot 3x – cot 3x cot x :

= \(\\\cot x \;\cot 2x – \cot 3x\left ( \cot 2x + \cot x \right )\\\)

= \(\\\cot x \;\cot 2x – \cot\left ( 2x + x \right )\left ( \cot 2x + \cot x \right )\\\)

= \(\\\cot x \;\cot 2x – \left [ \frac{\cot 2x \;\cot x – 1}{\cot x + \cot 2x} \right ]\left ( \cot 2x + \cot x \right )\\\)

= \(\\\left [\cot \left ( A + B \right ) = \frac{\cot A \; \cot B – 1}{\cot A + \cot B} \right ]\\\)

= \(\\\cot x \;\cot 2x – \left ( \cot 2x \cot x – 1 \right )\) = 1

= R.H.S

 

 

Q.23: Prove:

\(\tan 4x = \frac{4 \tan x \left ( 1 – \tan ^{2}x \right )}{1 – 6\tan ^{2} x + \tan ^{4} x}\)

 

Sol:

Since, tan 2A = \(\frac{2\tan A}{1 – \tan ^{2}A}\)

Now, taking L.H.S.

tan 4x :

= \(\tan 2\left (2x \right )\)

= \(\\\frac{2 \tan 2x}{1 – \tan ^{2}\left (2x \right )}\\\)

= \(\\\frac{2 \left ( \frac{2 \tan x}{1 – \tan ^{2}x} \right )}{1 – \left ( \frac{2 \tan x}{1 – \tan ^{2}x} \right )^{2}}\\\)

= \(\\\frac{\left ( \frac{4 \tan x}{1 – \tan ^{2}x} \right )}{1 – \left ( \frac{4 \tan ^{2}x}{\left (1 – \tan ^{2}x \right )^{2}} \right )}\\\)

=  \(\\\frac{\left (\frac{4\tan x}{1 – \tan ^{2}x} \right )}{\left [\frac{\left ( 1 – \tan ^{2}x \right )^{2} – 4 \tan ^{2}x}{\left ( 1 – \tan ^{2}x \right )^{2}} \right ]}\\\)

= \(\\\frac{4 \tan x \left ( 1 – \tan ^{2} x\right )}{\left ( 1 – \tan ^{2}x \right )^{2} – 4 \tan ^{2} x}\\\)

= \(\\\frac{4 \tan x \left ( 1 – \tan ^{2}x \right )}{1 + \tan ^{4}x – 2 \tan ^{2}x – 4 \tan ^{2}x}\\\)

= \(\\\frac{4 \tan x \left ( 1 – \tan ^{2}x \right )}{1 – 6 \tan ^{2}x + \tan ^{4}x }\)

= R.H.S.

 

 

Q.24: Prove:

\(\cos 4x = 1 – 8 \sin ^{2}x \;\cos ^{2}x\)

 

Sol:

Now, taking L.H.S.

cos 4x:

= \(\\\cos 2\left (2x \right )\\\)

= \(\\1 – 2 \sin ^{2} 2x \;\;\;\;\left [ \cos 2A = 1 – 2 \sin ^{2}A \right ]\\\)

= \(\\1 – 2 \left ( 2 \sin x \cos x \right )^{2} \;\;\;\;\;\left [ \sin 2A = 2 \sin A \cos A \right ]\\\)

= \(\\1 – 8 \sin ^{2}x \cos ^{2}x\\\)

= R.H.S.

 

 

Q.25: Prove:

\(\cos 6x = 32 \cos ^{6}x – 48 \cos ^{4}x + 18 \cos ^{2}x -1\)

 

Sol:

Now, taking L.H.S.

cos 3 (2x ):

= \(\\4 \cos ^{3}2x – 3 \cos 2x \;\;\;\;\; \left [ \cos 3A = 4\cos ^{3}A – 3 \cos A \right ]\\\)

= \(\\4 \left ( 2 \cos ^{2}x – 1 \right )^{3} – 3 \left ( 2 \cos ^{2}x -1 \right ) \;\;\;\;\; \left [ \cos 2x = 2 \cos ^{2}x -1 \right ]\\\)

= \(\\4 \left [ \left ( 2 \cos ^{2}x \right )^{3} – \left (1 \right )^{3} – 3 \left ( 2 \cos ^{2}x \right )^{2} + 3 \left ( 2 \cos ^{2}x \right )\right ] – 6 \cos ^{2}x + 3\\\)

= \(\\4 \left [ 8 \cos ^{6}x – 1 – 12\cos ^{4}x + 6 \cos ^{2}x \right ] – 6 \cos ^{2}x + 3\\\)

= \(\\32 \cos ^{6}x – 4 – 48 \cos ^{4}x + 24 \cos ^{2}x – 6 \cos ^{2}x + 3\\\)

= \(\\32 \cos ^{6}x – 48 \cos ^{4}x + 18 \cos ^{2}x -1\)

= R.H.S.

 

 

Exercise 3.4

 Q.1: Find general solutions and the principle solutions of the given equation: tan x = \(\sqrt{3}\)

 

Sol:

tan x = \(\sqrt{3}\)    [Given]

As we know that, \(\tan \frac{\pi }{3} = \sqrt{3}\)

And, \(\tan \frac{4\pi }{3}\) = \(\tan \left ( \pi + \frac{\pi }{3} \right )\) = \(\tan \left ( \frac{\pi }{3} \right )\) = \(\sqrt{3}\)

Therefore, the principle solutions are \(x = \frac{\pi }{3}\) and \(\frac{4\pi }{3}\)

Now, \(\tan x = \tan \frac{\pi }{3}\\\)

\(x = n\pi + \frac{\pi }{3}\), where n \(\in\) Z

Therefore, the general solution is \(x = n\pi + \frac{\pi }{3}\), where n \(\in\) Z.

 

 

Q.2: Find general solutions and the principle solutions of the given equation: sec x = 2

 

Sol:

sec x = 2   [Given]

As we know that, \(\sec \frac{\pi }{3} = 2\)

And, \(\sec \frac{5\pi }{3}\) = \(\sec \left (2\pi – \frac{\pi }{3} \right )\) = \(\sec \frac{\pi }{3}\) = 2

Therefore, the principle solutions are \(x = \frac{\pi }{3}\) and \(\frac{5\pi }{3}\).

Now, \(\sec x = \sec \frac{\pi }{3}\)

And, \(\cos x = \cos \frac{\pi }{3} \;\;\;\;\; \left [ \sec x = \frac{1}{\cos x} \right ]\\\)

\(x = 2n\pi \pm \frac{\pi }{3}\), where n \(\in\) Z

Therefore, the general solution is \(x = 2n\pi \pm \frac{\pi }{3}\), where n \(\in\) Z.

 

 

Q.3: Find general solutions and the principle solutions of the given equation:

cot = \( -\sqrt{3}\)

 

Sol:

cot = \(-\sqrt{3}\)          [Given]

As we know that, \(\cot \frac{\pi }{6} = \sqrt{3}\;\;\Rightarrow \;\;\cot \left (\pi – \frac{\pi }{6} \right ) = -\cot \frac{\pi }{6} = -\sqrt{3}\)

And, \(\cot \left ( 2\pi – \frac{\pi }{6} \right )\) = \(-\cot \frac{\pi }{6}\) = \(-\sqrt{3}\)

That is \(\cot \frac{5\pi }{6} = -\sqrt{3}\) and \(\cot \frac{11\pi }{6} = -\sqrt{3}\)

Therefore, the principle solutions are \(x = \frac{5\pi }{6}\) and \(\frac{11\pi }{6}\).

Now,\(\cot x = \cot \frac{5\pi }{6}\)

And, \(\tan x = \tan \frac{5\pi }{6} \;\;\;\;\;\left [ \cot x = \frac{1}{\tan x} \right ]\)

\(\\x = n\pi + \frac{5\pi }{6}\), where n \(\in\) Z

Therefore, the general solution is \(x = n\pi + \frac{5\pi }{6}\), where n \(\in\) Z.

 

 

Q.4: Find general solutions and the principle solutions of the given equation: cosec x = -2

 

Sol:

cosec x = -2     [Given]

As we know that, \(cosec \frac{\pi }{6} = 2\)

Hence, \(cosec \left (\pi + \frac{\pi }{6} \right )\) = \(-cosec \frac{\pi }{6}\) = -2

And, \(cosec \left (2\pi – \frac{\pi }{6} \right )\) = \(-cosec \frac{\pi }{6}\) = -2

That is \(cosec \frac{7\pi }{6} = -2\) and \(cosec \frac{11\pi }{6} = -2\).

Therefore, the principle solutions are \(x = \frac{7\pi }{6}\) and \(\frac{11\pi }{6}\).

Now,\(cosec \: x = cosec \frac{7\pi }{6}\)

And, \(\sin x = \sin \frac{7\pi }{6} \;\;\;\;\; \left [ cosec x = \frac{1}{\sin x} \right ]\)

\(\\x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}\), where n \(\in\) Z

Therefore, the general solution is \(x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}\), where n \(\in\) Z.

 

 

Q.5: Find the general solution of the given equation: cos 4x = cos 2x

Sol:

cos 4x = cos 2x           [Given]

i.e.  cos 4x – cos 2x = 0

\(-2\sin \left ( \frac{4x + 2x}{2} \right ) \sin \left ( \frac{ 4x – 2x }{2}\right ) = 0\)

Since, cos A – cos B = \(2\sin \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

(sin 3x)  (sin x) = 0

sin 3x = 0  or  sin x = 0

sin 3x = 0

\(3x = n\pi\\\)

\(x = \frac{n\pi}{3}\), where n \(\in\) Z

sin x = 0

\(\Rightarrow\)  \(x = n\pi\), where n \(\in\) Z

Q.6: Find the general solution of the given equation: cos 3x + cos x – cos 2x = 0

Sol:

cos 3x + cos x – cos 2x = 0        [Given]

\(2 \cos \left ( \frac{3x + x}{2} \right ) \cos \left ( \frac{3x – x}{2} \right ) – \cos 2x = 0\)

Since, cos A + cos B = \(2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

2cos 2x cos x – cos 2x = 0

cos 2x (2cos x – 1) = 0

cos 2x = 0  or  2cos x -1 = 0

cos 2x = 0

\(2x = \left ( 2n + 1 \right )\frac{\pi }{2}\), where n \(\in\) Z

\(\\x = \left ( 2n + 1 \right )\frac{\pi }{4}\), where n \(\in\) Z

2cos x -1 = 0

\(\cos x = \frac{1}{2}\\\)

\(\cos x = \cos \frac{\pi }{3}\), where n \(\in\) Z

\(x = 2n\pi \pm \frac{\pi }{3}\), where n \(\in\) Z

 

 

Q.7: Find the general solution of the given equation:  sin 2x + cos x = 0

 

Sol:

sin 2x + cos x = 0          [Given]

2sin x cos x + cos x = 0

cos x (2sin x + 1) = 0

cos x = 0   or   2sin x + 1 = 0

cos x = 0

\(\cos x = \left ( 2n + 1 \right )\frac{\pi }{2}\), where n \(\in\) Z

2sin x + 1 = 0

= \(-\sin \frac{\pi }{6}\) = \(\sin \left (\pi + \frac{\pi }{6} \right )\) = \(\sin \frac{7\pi }{6}\)

\(\Rightarrow\)  \(x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}\), where n \(\in\) Z

Therefore, the general solution is \(\\\left ( 2n + 1 \right )\frac{\pi }{2}\) or \(n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}\), where n \(\in\) Z.

 

 

Q.8: Find the general solution of the given equation:

\(\sec ^{2}2x = 1 – \tan 2x\)

 

Sol:

\(\sec ^{2}2x = 1 – \tan 2x\)           [Given]

\(1 + \tan ^{2}2x = 1 – \tan 2x\) \(\tan ^{2}2x + \tan 2x = 0\)

tan 2x ( tan 2x + 1) = 0

tan 2x = 0  or  tan 2x + 1 = 0

tan 2x = 0

tan 2x = tan 0

2x = n∏ + 0, where n \(\in\) Z

\(x = \frac{n\pi}{2}\), where n \(\in\) Z

tan 2x + 1 = 0

tan 2x = -1= \(-\tan \frac{\pi }{4}\)= \(\tan \left (\pi – \frac{\pi }{4} \right )\)= \(\tan \frac{3\pi }{4}\)

\(\Rightarrow\)  \(2x = n\pi + \frac{3\pi }{4}\), where n \(\in\) Z

\(\Rightarrow\)  \(x = \frac{n\pi}{2} + \frac{3\pi }{8}\), where n \(\in\) Z

Therefore, the general solution is \(\frac{n\pi}{2}\) or \(\frac{n\pi}{2} + \frac{3\pi }{8}\), where n \(\in\) Z.

 

 

Q.9: Find the general solution of the given equation:  sin x + sin 3x + sin 5x = 0

 

Sol:

sin x + sin 3x + sin 5x = 0              [Given]

(sin x + sin 5x) + sin 3x = 0

\(\left [ 2\sin \left ( \frac{x + 5x}{2} \right ) \cos \left ( \frac{x – 5x}{2} \right )\right ] + \sin 3x = 0\)

Since, sin A + sin B = \(2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

2sin 3x cos (-2x) + \sin 3x = 0

2sin 3x cos 2x + sin 3x = 0

sin 3x (2cos 2x + 1) = 0

sin 3x = 0 or  2cos 2x + 1 = 0

Now,

sin 3x = 0

\(3x = n\pi\), where n \(\in\) Z

\(x = \frac{n\pi }{3}\), where n \(\in\) Z

2cos 2x + 1 = 0

\(\cos 2x = -\frac{1}{2}\) = \(-\cos \frac{\pi }{3}\) = \(\cos \left (\pi – \frac{\pi }{3} \right )\) = \(\cos \frac{2\pi }{3}\)

\(2x = 2n\pi \pm \frac{2\pi }{3}\), where n \(\in\) Z

\(x = n\pi \pm \frac{\pi }{3}\), where n \(\in\) Z

Therefore, the general solution is \(\frac{n\pi }{3}\) or \(n\pi \pm \frac{\pi }{3}\), where n \(\in\) Z.

 

 

Miscellaneous Exercise

 Q.1: Prove that:

\(2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + \cos \frac{3\pi }{13}\; + \cos \frac{5\pi }{13} = 0\)

 

Sol:

Taking L.H.S.

\(2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + \cos \frac{3\pi }{13}\; + \cos \frac{5\pi }{13}\\\) :

= \(\\2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + 2\cos \left (\frac{\frac{3\pi }{13} + \frac{5\pi }{13}}{2} \right )\; \cos \left (\frac{\frac{3\pi }{13} – \frac{5\pi }{13}}{2} \right )\\\)

Since, \(\\\boldsymbol{\left [\cos x + \cos y = 2\cos \left ( \frac{x + y}{2} \right )\cos \left ( \frac{x – y}{2} \right ) \right ]}\\\)

= \(\\2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + 2\cos \frac{4\pi }{13} \; \cos \left (\frac{-\pi }{13} \right )\\\)

= \(\\2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + 2\cos \frac{4\pi }{13} \; \cos \frac{\pi }{13}\\\)

= \(\\2\cos \frac{\pi }{13}\; \left [\cos \frac{9\pi }{13}\; + \cos \frac{4\pi }{13} \right ]\\\)

= \(\\2\cos \frac{\pi }{13}\; \left [2\cos \left ( \frac{\frac{9\pi }{13} + \frac{4\pi }{13}}{2} \right )\; \cos \left ( \frac{\frac{9\pi }{13} – \frac{4\pi }{13}}{2} \right ) \right ]\\\)

= \(\\2\cos \frac{\pi }{13}\; \left [ 2\cos \frac{\pi }{2} \; \cos \frac{5\pi }{26}\right ]\\\)

= \(\\2\cos \frac{\pi }{13} \times 2 \times 0 \times \cos \frac{5\pi }{26}\)

= 0

= R.H.S.

Therefore, \(\boldsymbol{2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + \cos \frac{3\pi }{13}\; + \cos \frac{5\pi }{13} = 0}\)

 

 

Q.2: Prove that:

\(\left ( \sin 3x + \sin x \right )\sin x + \left ( \cos 3x – \cos x \right )\cos x = 0\)

 

Sol:

Taking L.H.S.

\(\\\left ( \sin 3x + \sin x \right )\sin x + \left ( \cos 3x – \cos x \right )\cos x\\\) :

= \(\\\sin 3x\; \sin x + \sin ^{2}x + \cos 3x \;\cos x – \cos ^{2}x\\\)

= \(\\\cos 3x \;\cos x + \sin 3x\; \sin x – \left (\cos ^{2}x – \sin ^{2}x \right )\\\)

= \(\\\cos \left ( 3x – x \right ) – \cos 2x\\\)

Since, \(\\\boldsymbol{\left [ \cos \left ( A -B \right ) = \cos A\; \cos B + \sin A \; \sin B \right ]}\\\)

= \(\\\cos 2x – \cos 2x\)

= 0

= R.H.S.

Therefore, \(\boldsymbol{\left ( \sin 3x + \sin x \right )\sin x + \left ( \cos 3x – \cos x \right )\cos x = 0}\)

 

 

Q-3: Prove that:

\(\left (\cos x + \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2} = 4\cos ^{2}\frac{x + y}{2}\)

 

Sol:

Taking L.H.S.

\(\left (\cos x + \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2}\\\) :

= \(\\\cos ^{2}x + \cos ^{2}y + 2\cos x\; \cos y + \sin ^{2}x + \sin ^{2}y – 2\sin x\; \sin y\\\)

= \(\\\left (\cos ^{2}x + \sin ^{2}x \right ) + \left (\cos ^{2}y + \sin ^{2}y \right ) + 2 \left (\cos x\; \cos y – \sin x\; \sin y \right )\\\)

= \(\\1 + 1 + 2\cos \left ( x +y \right )\\\)

Since, \(\\\boldsymbol{\left [ \cos \left ( A + B \right ) = \cos A\; \cos B – \sin A \; \sin B \right]}\\\)

= \(\\2 + 2\cos \left ( x +y \right )\\\)

= \(\\2\left [1 + \cos \left ( x +y \right ) \right ]\\\)

= \(\\2\left [1 + 2\cos ^{2}\left ( \frac{x + y }{2}\right ) – 1 \right ]\\\)

Since, \(\\\boldsymbol{\left [\cos 2A = 2\cos ^{2}A – 1 \right ]}\\\)

= \(\\4\cos ^{2}\left ( \frac{x + y}{2} \right )\\\)

= R.H.S.

Therefore, \(\boldsymbol{\left (\cos x + \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2} = 4\cos ^{2}\frac{x + y}{2}}\)

 

 

Q-4: Prove that:

\(\left ( \cos x – \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2} = 4\sin ^{2}\frac{x – y}{2}\)

 

Sol:

Taking L.H.S.

\(\left ( \cos x – \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2}\\\) :

= \(\\\cos ^{2}x + \cos ^{2}y – 2\cos x \; \cos y + \sin ^{2}x + \sin ^{2}y – 2\sin x\; \sin y\\\)

= \(\\\left (\cos ^{2}x + \sin ^{2}x \right ) + \left (\cos ^{2}y + \sin ^{2}y \right ) – 2 \left (\cos x\; \cos y – \sin x\; \sin y \right )\\\)

= \(\\1 + 1 – 2\left [ \cos \left ( x – y \right ) \right ]\\\)

Since, \(\\\boldsymbol{\left [ \cos \left ( A -B \right ) = \cos A\; \cos B + \sin A \; \sin B \right ]}\\\)

= \(\\2\left [ 1 – \cos \left ( x – y \right ) \right ]\\\)

= \(\\2\left [ 1 – \left \{ 1 – 2 \sin ^{2}\left ( \frac{x – y}{2} \right ) \right \} \right ]\\\)

Since, \(\\\boldsymbol{\left [\cos 2A = 1 – 2\sin ^{2}A \right ]}\\\)

= \(\\4\sin ^{2}\frac{x – y}{2}\)

= R.H.S.

Therefore, \(\boldsymbol{\left ( \cos x – \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2} = 4\sin ^{2}\frac{x – y}{2}}\)

 

 

Q-5: Prove that:

\(\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x \; \cos 2x \; \cos 4x\)

 

Sol:

Taking L.H.S.

\(\sin x + \sin 3x + \sin 5x + \sin 7x\\\) :

= \(\\\left (\sin x + \sin 5x \right ) + \left (\sin 3x + \sin 7x \right )\\\)

= \(\\2 \sin \left ( \frac{x + 5x}{2} \right )\; \cos \left ( \frac{x – 5x}{2} \right ) + 2 \sin \left ( \frac{3x + 7x}{2} \right )\; \cos \left ( \frac{3x + 7x}{2} \right )\\\)

Since, \(\\\boldsymbol{\left [\sin A + \sin B = 2\sin \left ( \frac{A +B}{2} \right ).\cos \left ( \frac{A – B}{2} \right ) \right ]}\\\)

= \(\\2\sin 3x \cos \left ( -2x \right ) + 2\sin 5x \cos \left ( -2x \right )\\\)

= \(\\2\sin 3x \cos 2x + 2\sin 5x \cos 2x\\\)

= \(\\2 \cos 2x \left [ \sin 3x + \sin 5x\right ]\\\)

= \(\\2 \cos 2x \left [2 \sin \left ( \frac{3x + 5x}{2} \right ). \cos \left ( \frac{3x – 5x}{2} \right )\right ]\\\)

= \(\\2 \cos 2x \left [2 \sin 4x. \cos \left ( -x \right )\right ]\\\)

= \(\\4 \cos 2x\; \sin 4x \; \cos x\)

= R.H.S.

Therefore, \(\boldsymbol{\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x \; \cos 2x \; \cos 4x}\)

 

 

Q-6: Prove that:

\(\frac{\left ( \sin 7x + \sin 5x \right ) + \left ( \sin 9x + \sin 3x \right )}{\left ( \cos 7x + \cos 5x \right ) + \left ( \cos 9x + \cos 3x \right )} = \tan 6x\)

 

Sol:

Since, \(\boldsymbol{\sin A + \sin B = 2\sin \left ( \frac{A + B}{2} \right ). \cos \left ( \frac{A -B}{2} \right )}\\\)

And, \(\\\boldsymbol{\cos A + \cos B = 2\cos \left ( \frac{A + B}{2} \right ). \cos \left ( \frac{A -B}{2} \right )}\)

Taking L.H.S.

\(\\\frac{\left ( \sin 7x + \sin 5x \right ) + \left ( \sin 9x + \sin 3x \right )}{\left ( \cos 7x + \cos 5x \right ) + \left ( \cos 9x + \cos 3x \right )}\\\) :

= \(\\\frac{\left [2\sin \left ( \frac{7x + 5x}{2} \right ). \cos \left ( \frac{7x – 5x}{2} \right ) \right ] + \left [ 2 \sin \left ( \frac{9x + 3x}{2}\right ). \cos \left ( \frac{9x – 3x}{2}\right ) \right ]}{\left [2\cos \left ( \frac{7x + 5x}{2} \right ). \cos \left ( \frac{7x – 5x}{2} \right ) \right ] + \left [ 2 \cos \left ( \frac{9x + 3x}{2}\right ). \cos \left ( \frac{9x – 3x}{2}\right ) \right ]}\\\)

= \(\\\frac{\left [2\sin 6x. \cos x \right ] + \left [ 2 \sin 6x.\cos 3x \right ]}{\left [2\cos 6x.\cos x \right ] + \left [ 2\cos 6x. \cos 3x \right ]}\\\)

= \(\\\frac{2\sin 6x \left [\cos x + \cos 3x \right ]}{2\cos 6x \left [\cos x + \cos 3x \right ] }\\\)

= \(\tan 6x\)

= R.H.S.

Therefore, \(\boldsymbol{\frac{\left ( \sin 7x + \sin 5x \right ) + \left ( \sin 9x + \sin 3x \right )}{\left ( \cos 7x + \cos 5x \right ) + \left ( \cos 9x + \cos 3x \right )} = \tan 6x}\)

 

 

Q-7: Show that: \(\sin 3y + \sin 2y – \sin y = 4\sin y \cos\frac{y}{2} \cos\frac{3y}{2}\)

 

Sol:

Here, L.H.S = \(\sin 3y + \sin 2y\; – \sin y\\\)

= \(\;\\\sin 3y + (\sin 2y \;- \sin y) = \sin 3y + \left [ 2\cos\left ( \frac{2y + y}{2} \right ) \sin\left ( \frac{2y – y}{2} \right ) \right ]\\\)

Since,  \(\boldsymbol{(\sin x – \sin y) = \left [ 2\cos\left ( \frac{x + y}{2} \right ) \sin\left ( \frac{x – y}{2} \right ) \right ]}\)

= \(\\\sin 3y + \left [ 2\cos\left ( \frac{3y}{2} \right ) \sin\left ( \frac{y}{2} \right ) \right ]\\\)

=\( \left [ 2\sin\left ( \frac{3y}{2} \right ) \cos\left ( \frac{3y}{2} \right ) \right ] + \left [ 2\cos\left ( \frac{3y}{2} \right ) \sin\left ( \frac{y}{2} \right ) \right ]\\\)

Since, sin 2x = 2 sin x cos x

=\(\\2\cos\frac{3y}{2}\cdot [ \sin\ \frac{3y}{2} + \sin \frac{y}{2}]\\\)

=\(2\cos\frac{3y}{2}\cdot [ \sin\ \frac{(\frac{3y}{2} + \frac{y}{2})}{2}][ \cos\ \frac{(\frac{3y}{2} – \frac{y}{2})}{2}]\)

Since, sin x + sin y = \(\boldsymbol{\\2\sin\left ( \frac{x + y}{2} \right )\cos\left ( \frac{x – y}{2} \right )}\)

\(\\= 2\cos\left ( \frac{3y}{2} \right )2\sin y \cos\left ( \frac{y}{2} \right )\) \(\\= 4\sin y\cos\left ( \frac{y}{2} \right ) \cos\left ( \frac{3y}{2} \right )\)

= R.H.S

 

 

Q-8: The value of \(\tan y = -\frac{4}{3}\) where y in in 2nd quadrant then find out the values of \(\sin \frac{y}{2}, \cos \frac{y}{2} \;and\; \tan\frac{y}{2}\).

 

Sol:

Here, y is in 2nd quadrant.

So, \(\frac{\pi}{2} < y < \pi \\ \\ \Rightarrow \frac{\pi}{4} < \frac{y}{2} < \frac{\pi}{2}\)

Thus, \(\sin \frac{y}{2}, \cos \frac{y}{2} \;and\; \tan\frac{y}{2}\) lies in 1st quadrant.

Now,

\(\tan y = -\frac{4}{3}\) \(\sec^{2} y = 1 + \tan^{2} y = 1 + (\frac{4}{3})^{2} = 1 + \frac{16}{9} = \frac{25}{9}\\\)

So, \(\cos^{2} y = \frac{9}{25}\)

\(\Rightarrow \cos y = \pm \frac{3}{5}\)

As y is in 2nd quadrant, cos y is negative.

\(\cos y = \frac{-3}{5}\)

So, cos y = \(2\cos^{2} \frac{y}{2} – 1\)

\(\Rightarrow \frac{-3}{5} = 2\cos^{2} \frac{y}{2} – 1 \\ \\ \Rightarrow 2\cos^{2} \frac{y}{2} = 1 – \frac{-3}{5} \\ \\ \Rightarrow 2\cos^{2} \frac{y}{2} = \frac{2}{5} \\ \\ \Rightarrow 2\cos^{2} \frac{y}{2} = \frac{1}{5} \\ \\ \Rightarrow 2\cos^{2} \frac{y}{2} = \frac{3}{\sqrt{5}}\)  [Since, \(\boldsymbol{\cos\frac{y}{2}}\) is positive]

\(\Rightarrow \cos\frac{y}{2} = \frac{\sqrt{5}}{5} \\ \\ \sin^{2} \frac{y}{2} + \cos^{2} \frac{y}{2} = 1 \\ \\ \Rightarrow \sin^{2} \frac{y}{2} + \cos^{2} \frac{1}{\sqrt{5}} = 1 \\ \\ \Rightarrow \sin^{2} \frac{y}{2} = 1 – \frac{1}{5} = \frac{4}{5} \\ \\ \Rightarrow \sin^{2} \frac{y}{2} = \frac{2}{\sqrt{5}}\)  [Since,  \(\boldsymbol{\sin\frac{y}{2}}\) is positive]

\(\Rightarrow \sin^{2} \frac{y}{2} = \frac{2\sqrt{5}}{5} \\ \\ \tan \frac{y}{2} = \frac{\sin\frac{y}{2}}{\cos\frac{y}{2}} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = 2\)

 

 

Q-9: The value of \(\cos y = -\frac{1}{3}\) where y in in 3rd quadrant then find out the values of \(\sin \frac{y}{2}, \cos \frac{y}{2} \;and\; \tan\frac{y}{2}\).

 

Sol:

Here, y is in 3rd  quadrant.

So, \(\pi < y < \frac{3\pi}{2} \\ \\ \Rightarrow \frac{\pi}{2} < \frac{y}{2} < \frac{3\pi}{4}\)

Thus, \(\cos \frac{y}{2} \;and\; \tan\frac{y}{2}\) are negative and \(\sin \frac{y}{2}\) is positive.

Now,  \(\cos y = -\frac{1}{3}\)  [Given]

\(\cos y = 1 – 2\sin^{2}\frac{y}{2} \Rightarrow \sin^{2}\frac{y}{2} = \frac{1 – \cos y}{2} \Rightarrow \sin^{2}\frac{y}{2} = \frac{1 – \frac{-1}{3}}{2} = \frac{1 + \frac{1}{3}}{2} = \frac{2}{3}\\\)

\(\\\Rightarrow \boldsymbol{\sin \frac{y}{2}} = \frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}\\\)   [Since, \(\sin\frac{y}{2}\) is positive]

Now, \(\cos y = 2\cos^{2}\frac{y}{2} – 1\)

\(2\cos^{2}\frac{y}{2} = \frac{1 + \cos y}{2} = \frac{1 – \frac{1}{3}}{2} = \frac{1}{3}\\\)

\(\\\Rightarrow\)   \(\boldsymbol{\cos \frac{y}{2}}= \frac{-1}{\sqrt{3}} = \frac{-1}{\sqrt{3}}\times \frac{-\sqrt{3}}{3} = \frac{-1}{\sqrt{3}}\)

Therefore, \(\boldsymbol{\tan\frac{y}{2}} = \frac{\sin\frac{y}{2}}{\cos\frac{y}{2}} = \frac{\frac{\sqrt{2}}{\sqrt{3}}}{\frac{-1}{\sqrt{3}}} = -\sqrt{2}\)

 

 

Q-10: The value of \(\sin y = \frac{1}{4}\) where y in in 2nd quadrant then find out the values of \(\sin \frac{y}{2}, \cos \frac{y}{2} \;and\; \tan\frac{y}{2}\).

 

Sol:

Here, y is in 2nd quadrant.

So, \(\frac{\pi}{2} < y < \pi \\ \\ \Rightarrow \frac{\pi}{4} < \frac{y}{2} < \frac{\pi}{2}\)

Thus, \(\sin \frac{y}{2}, \cos \frac{y}{2} \;and\; \tan\frac{y}{2}\) lies in 1st quadrant.

Now, \(\sin y = \frac{1}{4}\)

\(\\\cos^{2} y = 1 – 2\sin^{2} y = 1 – (\frac{1}{4})^{2} = 1 – \frac{1}{16} = \frac{15}{16} \\ \\ \Rightarrow \cos y = -\frac{\sqrt{15}}{4}\)  [Since, \(\cos y\) is negative]

\(\sin^{2} \frac{y}{2} = \frac{1 – (\frac{-\sqrt{15}}{4})}{2} = \frac{4 + \sqrt{15}}{8}\\\)

\(\\\Rightarrow \)   \(\boldsymbol{\sin \frac{y}{2}} = \sqrt{\frac{4 + \sqrt{15}}{8}}\)  [ Since, \(\sin\frac{y}{2}\\\) is positive ]

=\(\\\sqrt{\frac{4 + \sqrt{15}}{8}\times \frac{2}{2}}= \sqrt{\frac{8 + 2\sqrt{15}}{16}} =\boldsymbol{\frac{\sqrt{8 + 2\sqrt{15}}}{4}}\\\)

\(\\\cos^{2} y = \frac{1 + \cos y}{2} = \frac{1 + (-\frac{\sqrt{15}}{4})}{2} = \frac{4 – \sqrt{15}}{8}\\\)

Therefore, \(\boldsymbol{\cos \frac{y}{2}}= \sqrt{\frac{4 – \sqrt{15}}{8}}\) = \(\sqrt{\frac{4 – \sqrt{15}}{8}\times \frac{2}{2}}\\\)

= \(\sqrt{\frac{8 – 2\sqrt{15}}{16}}=\boldsymbol{\frac{\sqrt{8 – 2\sqrt{15}}}{4}}\)

Now, \(\boldsymbol{tan \frac{y}{2}} = \frac{\sin \frac{y}{2}}{\cos \frac{y}{2}} = \frac{\left ( \frac{\sqrt {8\; + \;2\sqrt{15}}}{4} \right )}{\left ( \frac{\sqrt {8\; -\; 2\sqrt{15}}}{4} \right )}\)

= \(\\\frac{\sqrt{8 + 2\sqrt{15}}}{\sqrt{8 – 2\sqrt{15}}} = \frac{\sqrt{8 + 2\sqrt{15}}}{\sqrt{8 – 2\sqrt{15}}}\times \frac{\sqrt{8 + 2\sqrt{15}}}{\sqrt{8 + 2\sqrt{15}}}\\\\\)

= \(\\\frac{\sqrt{(8 + 2\sqrt{15})^{2}}}{\sqrt{64\; -\; 60}} = \frac{8 + 2\sqrt{15}}{2} =\boldsymbol{4 + \sqrt{15}}\)

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