# 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?

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 = $$\frac{22}{7}$$]

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.

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 $$\sin \left ( -\frac{11\pi }{3} \right )$$.

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:

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

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

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

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