Trigonometry

 

Trigonometric Ratios

\(\sin \theta =\frac{AB}{AC}=\frac{opp}{hyp}\) \(\cos \theta =\frac{BC}{AC}=\frac{adj}{hyp}\) \(\tan \theta =\frac{AB}{BC}=\frac{opp}{adj}\) \(\cot\theta =\frac{BC}{AB}\) \(\sec\theta =\frac{AC}{BC}\)

\(\cos ec\theta =\frac{AC}{AB}\)

Trigonometric Circular Function

\(<AOP=\frac{arcAP}{r}=\frac{l}{r}\)

Cos \(\theta\)= x

Sin \(\theta\) = y

\(\tan\theta=\frac{x}{y}\)

Graphs of T Ratio

  • Sine

Y= Sinx

Domain R

Range (-1,1)

  • Cosine

y = cos x

  • Tangent

y = tan x

 

 

 

  • Co – tangent

y = cot x

Domain

\(x\sum{R-(n\pi )}\)

Range (\(-\infty ,\infty\))

  • Secant
\(y=\sec x,\)

\(Domain\,\,\,\,\,R-\left\{ (2x+1)\,\frac{\pi }{2} \right\}\) \(Range\,(-\infty -1)\,u\,(1,\infty )\)
  • Cosecant

y = cosecx

.\(Domain\,R-\{n\pi \}\) \(Range:\,(-\infty -1)\,u\,(1,\infty )\)

Watch this Video for More Reference


Trignometric Identities

\({{\sin }^{2 }\theta}+{{\cos }^{2 }\theta}=1\) \(1+{{\tan }^{2 }\theta}=\,{{\sec }^{2 }\theta}\) \(1+{{\cot }^{2 }\theta}=\cos e{{c}^{2 }\theta}\) \(\sin ^{4}\theta +\cos^{4}=1-2\sin ^{2 }\theta\cos ^{2 }\theta\) \({{\sin }^{6}}\theta +{{\cos }^{6}}=1-3{{\sin }^{2 }\theta}{{\cos }^{2 }\theta}\)

Remember:

\(\sec \theta -\tan \theta =\frac{1}{\sec \theta +\tan \theta }\) \(\cos ec\theta -\cot \theta =\frac{1}{\cos ec\theta +\cot \theta }\)

Sign Of T-Ratios:

T-Ratio At Some Standard Angles:

T-Ratio At Some Specific Angles:

T-ratio

\(7\frac{1{}^\circ }{2}\) 15o \(22\frac{1{}^\circ }{2}\) 18o
\(\sin\theta\) \(\frac{\sqrt{4-\sqrt{2}-\sqrt{6}}}{2\sqrt{2}}\) \(\frac{\sqrt{3}-1}{2\sqrt{2}}\) \(\frac{1}{2}\sqrt{2-\sqrt{2}}\) \(\frac{\sqrt{5-1}}{4}\)
\(\cos\theta\) \(\frac{\sqrt{4+\sqrt{2}+\sqrt{6}}}{2\sqrt{2}}\) \(\frac{\sqrt{3}+1}{2\sqrt{2}}\) \(\frac{1}{2}\sqrt{2+\sqrt{2}}\) \(\frac{1}{4}\sqrt{10+2\sqrt{5}}\)
\(\tan\theta\) \(\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{2}-1 \right)\) \(2-\sqrt{3}\) \(\sqrt{2-1}\) \(\frac{\sqrt{25+10\sqrt{15}}}{5}\)

Maximum and Minimum Value Of Trigonometric Expressions

  • Maximum value of \(a\cos \theta \pm b\sin \theta =\sqrt{{{a}^{2}}+{{b}^{2}}}\)
  • Maximum value of \(a\cos \theta \pm b\sin \theta =-\sqrt{{{a}^{2}}+{{b}^{2}}}\)
  • Maximum value of \(a\cos \theta \pm b\sin \theta +c=c+\sqrt{{{a}^{2}}+{{b}^{2}}}\)
  • Maximum value of \(a\cos \theta \pm b\sin \theta +c=c-\sqrt{{{a}^{2}}+{{b}^{2}}}\)

Problems on Trignometry

Example:

\({{\sec }^{4}}x-{{{cosec}}^{4}}x-2{{\sec }^{2}}x+2\cos e{{c}^{2}}x=\frac{15}{4}\)

Find tan x.

Solution:

\(\left( {{\sec }^{4}}x-2{{\sec }^{2}}x \right)-\left( \cos e{{c}^{4}}x-2\cos e{{c}^{2}}x \right)=\frac{15}{4}\)

Adding and subtracting

\(\left( {{\sec }^{4}}x-2{{\sec }^{2}}x+1 \right)-\left( \cos e{{c}^{4}}x-2\cos e{{c}^{2}}x+1 \right)=\frac{15}{4}\) \({{\left( {{\sec }^{2}}x-1 \right)}^{2}}-{{\left( \cos e{{c}^{2}}x-1 \right)}^{2}}=\frac{15}{4}\) \({{\left( {{\tan }^{2}}x \right)}^{2}}-{{\left( {{\cot }^{2}}x \right)}^{2}}=\frac{15}{4}\) \({{\tan }^{4}}x-\frac{1}{{{\cot }^{4}}x}=\frac{15}{4}\) \({{\tan }^{4}}x-\frac{1}{{{\cot }^{4}}x}=4-\frac{1}{4}\)

On comparing,

\({{\tan }^{4}}x=4\) \({{\tan }^{4}}x=2\) \({{\tan }^{4}}x=\pm \sqrt{2}\)

Example:

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

\({{\cos }^{8}}x+2{{\cos }^{6}}x+{{\cos }^{4}}x=?\)

Solution:

\({{\cos }^{2}}x=\sin x\) \({{\sin }^{4}}x+2{{\sin }^{3}}x+{{\sin }^{2}}x\) \({{\left( {{\sin }^{2}}x+\sin x \right)}^{2}}\) \({{\left( 1 \right)}^{2}}=1\)

Example:

If \(\sin \theta ,\cos \theta ,\tan \theta\) are in G.P.

Then \({{\cos }^{9}}\theta +{{\cos }^{6}}\theta +3{{\cos }^{5}}\theta -1=?\)

Solution:

\(\sin \theta .tan\theta =co{{s}^{2}}\theta\) \(\frac{\sin \theta }{\cos \theta }=co{{s}^{2}}\theta\) \({{\sin }^{2}}\theta ={{\cos }^{3}}\theta\) \({{\left( {{\cos }^{3}}\theta \right)}^{3}}+{{\left( {{\cos }^{3}}\theta \right)}^{2}}+3{{\cos }^{3}}\theta -{{\cos }^{2}}\theta -1\) \({{\sin }^{6}}\theta +{{\sin }^{4}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta -1\) \({{\sin }^{6}}\theta +{{\sin }^{4}}\theta -\left( 1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)\) \({{\sin }^{6}}\theta +{{\sin }^{4}}\theta -\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)\) \({{\sin }^{4}}\theta -{{\cos }^{6}}\theta\) \({{\sin }^{4}}\theta -{{\left( {{\cos }^{3}}\theta \right)}^{2}}\) \({{\sin }^{4}}\theta -{{\sin }^{4}}\theta =0\)

Example:

If \(\sec x-\tan x=P\) then sec x = ?

Solution:

\(\sec x-\tan x=\frac{1}{\sec x+\tan x}\) \(\sec x\;+\;tan x=\frac{1}{P}\) \(\sec x\;-\;tan x=P\) \(\overline{2\sec =P+\frac{1}{P}}\) \(\sec x=\frac{{{P}^{2}}+1}{2P}\)

Example:

Prove that \(\sin \left( -420 \right).\cos \left( 390 \right)+\cos \left( -660 \right)\sin \left( 330 \right)=-1\)

Solution:

\(-\sin \left( 420 \right)\cos \left( 390 \right)+\cos \left( 660 \right)\sin 330\) \(-\sin \left( 2\pi +60 \right)\cos \left( 2\pi +30 \right)+\cos \left( 4\pi -60 \right)\sin \left( 2\pi -30 \right)\) \(-\sin 60\cos 30+\cos 60\left( -\sin 30 \right)\) \(-\frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}-\frac{1}{2}.\frac{1}{2}\) \(-\frac{3}{4}-\frac{1}{4}=-\frac{4}{4}=-1.\)

Example:

  1. \(\sin 1{}^\circ >0T/F?\)

sin 57o70

Hence True.

  1. \(\cos 1{}^\circ >0T/F?\)

cos 57o70

Hence True.

Trignometric Ratios of Compound Angles

The algebraic sum of two or more angles are generally called compound angles and angles are known as constituent angles.

Some Important Results:

  1. \(\sin \left( A\pm B \right)=\sin A\cos B\pm \cos \,A\,sin\,B\)
  2. \(\cos \left( A\pm B \right)=\cos A\cos B\mp \sin \,A\,sin\,B\)
  3. \(\tan \left( A\pm B \right)=\frac{\tan A\pm \tan B}{1\tan A\tan \,B}\)
  4. \(\cot \left( A\pm B \right)=\frac{\cot A\cot B\mp 1}{\cot B\pm \cot A}\)
  5. \(\sin \left( A+B \right)\sin \left( A-B \right)={{\sin }^{2}}A-{{\sin }^{2}}B={{\cos }^{2}}B-{{\cos }^{2}}A\)
  6. \(\cos \left( A+B \right)\cos \left( A-B \right)={{\cos }^{2}}A-{{\sin }^{2}}B={{\cos }^{2}}B-{{\sin }^{2}}A\)
  7. \(\sin \left( A+B+C \right)=\sin A\cos B\cos C+\sin B\cos A\cos C+\sin C\cos A\cos B-\sin A\sin B\sin C\)
  8. \(\cos \left( A+B+C \right)=\cos A\cos B\cos C-\cos A\sin B\sin C-\cos B\sin A\sin C-\cos C\sin A\sin B\)
  9. \(\tan \left( A+B+C \right)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\)

Remember:

If A+B+C = 0 then

\(\tan A+\tan B+\tan C=\tan A\tan B\tan C\)

Example:

If \(\sin \alpha =\frac{3}{5}\And \cos \beta =\frac{9}{41}\alpha ,\beta \in\) I quadrant

find \(\sin \left( \alpha -\beta \right),\cos \left( \alpha -\beta \right)\)

Solution:

\(\sin \alpha =\frac{3}{5}\) \({{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\) \({{\cos }^{2}}\alpha =1\frac{-9}{25}=\frac{16}{25}\) \(\cos \alpha =\frac{4}{5},\frac{-4}{5}\)as in first quadrant.

\(\cos \beta =\frac{9}{41}\) \({{\cos }^{2}}\beta +{{\sin }^{2}}\beta =1\) \({{\sin }^{2}}\beta =1\frac{-81}{1681}=\frac{1600}{1681}\) \(\sin \beta =\frac{40}{41}\) \(\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta\) \(=\frac{3}{5}\times \frac{9}{41}-\frac{4}{5}\times \frac{40}{41}=\frac{27-160}{205}\) \(=\frac{-133}{205}.\)

Example:

Find value of \(\frac{\sin \left( B-C \right)}{\cos B\cos C}+\frac{\sin \left( C-A \right)}{\cos C\cos A}+\frac{\sin \left( A-B \right)}{\cos A\cos B}=?\)

Solution:

\(\frac{\sin B\cos C-\cos B\sin C}{\cos B\cos C}+\frac{\sin C\cos A-\cos C\sin A}{\cos C\cos A}+\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}\) \(tan \;B\;-\;tan \;C\;+\;tan C\;-\;tan A\;+\;tan A\;-\;tan\; B = 0\)

Example:

Find value of \(\frac{{{\tan }^{2}}\left( 2\theta \right)-{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\left( 2\theta \right){{\tan }^{2}}\theta }=?\)

Solution:

\(\frac{\left( {{\tan }^{2}}\left( 2\theta \right)+{{\tan }^{2}}\theta \right)\left( \tan 2\theta -\tan \theta \right)}{\left( 1+\tan 2\theta \tan \theta \right)\left( 1-\tan 2\theta \tan \theta \right)}\) \(\tan 3\theta .\tan \theta \)

Example:

Prove that \(\tan 8\theta -\tan 6\theta -\tan 2\theta =\tan 8\theta .\tan 6\theta .\tan 2\theta .\) \(\tan 8\theta -\tan 2\theta \tan 6\theta \tan 8\theta =\tan 2\theta +\tan 6\theta\) \(\tan 8\theta -\tan 6\theta -\tan \theta =\tan 2\theta \tan 6\theta \tan 8\theta .\)

Transformation Formulae:

  1. \(2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)\)
  2. \(2\sin B\cos A=\sin \left( A+B \right)-\sin \left( A-B \right)\)
  3. \(2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\)
  4. \(2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)\)
  5. \(\sin C+\sin D=2\sin \left( \frac{C+1}{2} \right)\cos \left( \frac{C-D}{2} \right)\)
  6. \(\sin C-\sin D=2\cos \left( \frac{C+D}{2} \right)\sin \left( \frac{C-D}{2} \right)\)
  7. \(\cos C-\cos D=2\cos \left( \frac{C+D}{2} \right)\cos \left( \frac{C-D}{2} \right)\)
  8. \(\cos C-\cos D=2\sin \left( \frac{C+D}{2} \right)\sin \left( \frac{D-C}{2} \right)\)

T-Ratio Of Multiple Angles

  1. \(\sin 2A=2\sin A\cos A=\frac{2\tan A}{1+{{\tan }^{2}}A}\)
  2. \(\cos 2A=2{{\cos }^{2}}A-1=1-2{{\sin }^{2}}A=\frac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\)
  3. \(\tan 2A=\frac{2\tan A}{1-{{\tan }^{2}}A}\)
  4. \(\sin 3A=3\sin A-4{{\sin }^{3}}A\)
  5. \(\cos 3A=4{{\cos }^{3}}A-3\cos A\)
  6. \(\tan 3A=\frac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}\)

Remember:

\(1+\cos 2A=2{{\cos }^{2}}A\) \(1-\cos 2A=2{{\sin }^{2}}A\) \(\frac{1-\cos 2A}{1+\cos 2A}={{\tan }^{2}}A\)

Values Of T-Ratios At Some Useful Angles

  1. \(\sin 75{}^\circ =\frac{\sqrt{3}+1}{2\sqrt{2}}=\cos 15{}^\circ\)
  2. \(\cos 75{}^\circ =\frac{\sqrt{3}-1}{2\sqrt{2}}=\sin 15{}^\circ\)
  3. \(\tan 75{}^\circ =2+\sqrt{3}=\cot 15{}^\circ\)
  4. \(\cot 75{}^\circ =2-\sqrt{3}=\tan 15{}^\circ\)
  5. \(\sin 9{}^\circ =\frac{\sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}}}{4}=\cos 81{}^\circ\)
  6. \(\cos 9{}^\circ =\frac{\sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}}}{4}=\sin 81{}^\circ\)

Important Results:

\(\sin \theta \sin \left( 60-\theta \right)\sin \left( 60+\theta \right)=\frac{1}{4}\sin \theta\) \(\cos \theta \cos \left( 60-\theta \right)\cos \left( 60+\theta \right)=\frac{1}{4}\cos \theta\) \(\tan \theta \tan \left( 60-\theta \right)\tan \left( 60+\theta \right)=\tan 3\theta\)

Example:

If \(A+B=225{}^\circ\) \(\left( \frac{\cot A}{1+\cot A} \right)\left( \frac{\cot B}{1+\cot B} \right)=?\).

Solution:

\(A+B=225=180+45{}^\circ\) \(\tan \left( A+B \right)=\tan \left( \bar{\wedge }+45{}^\circ \right)\) \(\tan \left( A+B \right)=\tan 45{}^\circ\) \(A+B=\tan 45{}^\circ\) \(\left( \frac{\frac{1}{\tan A}}{1+\frac{1}{\tan A}} \right)\left( \frac{\frac{1}{\tan B}}{1+\frac{1}{\tan B}} \right)\) \(\frac{1}{\left( 1+\tan A \right)\left( 1+{tanB} \right)}=\frac{1}{1+\tan A\tan B+\tan B+\tan A}\)

= \(\frac{1}{\left( 1+\tan A \right)\left( 1+{tanB} \right)}=\frac{1}{1+\tan A\tan B+\tan B+\tan A}\) \(=\frac{1}{2}\)

Example:

Find value of \(\frac{\sin A\sin 2A+\sin 3A\sin 6A+\sin 4A\sin 13A}{2\sin A\cos 2A+sin3Acos6A+sin4Acos13A}=?\)

Solution:

\(\frac{\cos \left( -A \right)-{\cos \left( 3A \right)}+{\cos 3A}-{\cos 9A}+{\cos 9A}-\cos 17A}{{\sin 3A}+\sin \left( -A \right)+{\sin 9A}+{sin\left( -3A \right)}+\sin \left( 17A \right)+{\sin \left( -9A \right)}}\) \(\frac{\cos A-\cos 17A}{\sin 17A-\sin A}\) \(\frac{{2}\sin 9A{\sin 8A}}{{2}\cos 9A{\sin 8A}}\)

tan 9A

Example:

Prove that \(\cos A\sin \left( B-C \right)+\cos B\sin \left( C-A \right)+\cos C\sin \left( A-B \right)=0\)

Solution:

\(\sin {\left( A+B-C \right)}-\sin {\left( A-B+C \right)}+\sin {\left( B+C-A \right)}-\sin {\left( B-C+A \right)}+\sin {\left( C+A-B \right)}-\sin {\left( C-A+B \right)}\)

Example:

\({{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +{{\sin }^{2}}15{}^\circ +……….+{{\sin }^{2}}90{}^\circ =?\)

Solution:

\({{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +……….+{{\sin }^{2}}85{}^\circ +{{\sin }^{2}}90{}^\circ\) \({{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +……….+{{\cos }^{2}}5{}^\circ +{{\sin }^{2}}90{}^\circ\) \(\left( {{\sin }^{2}}5{}^\circ +{{\cos }^{2}}5{}^\circ \right)+……….+{{\sin }^{2}}90{}^\circ +{{\sin }^{2}}45{}^\circ\) \(8\times 1+1+\frac{1}{2}\) \(8+1+\frac{1}{2}=9\frac{1}{2}\)

Example:

Prove that \(\sin \alpha +sin\left( \alpha +\frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{4\pi }{3} \right)=0\)

Solution:

\(2\sin \left( \frac{2\alpha }{2}+\frac{4\pi }{6} \right)\cos \left( -\frac{4\pi }{6} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right)\) \(2\sin \left( \alpha +\frac{2\pi }{3} \right)\cos \left( \frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right)\) \(2\times \left( -\frac{1}{2} \right)\sin \left( \alpha +\frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right)\) \(-\sin \left( \alpha +\frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right)\)

=0

Example:

If \(a\sin \theta =b\sin \left( \theta +\frac{2\pi }{3} \right)=c\sin \left( \theta +\frac{4\pi }{3} \right)\) \(P.Tab+bc+ca=0\)

Solution:

\(a\sin \theta =b\sin \left( \theta +\frac{2\pi }{3} \right)=c\sin \left( \theta +\frac{4\pi }{3} \right)=k\) \(\sin \theta =\frac{k}{a}\) \(\sin \left( \theta +\frac{2\pi }{3} \right)=\frac{k}{b}\) \(\sin \left( \theta +\frac{4\pi }{3} \right)=\frac{k}{c}\)

We know

\(\sin \theta +\sin \left( \theta +\frac{2\pi }{3} \right)+\sin \left( \theta +\frac{4\pi }{3} \right)=0\) \(\frac{k}{a}+\frac{k}{b}+\frac{k}{c}=0\) \(\frac{k\left( bc+ac+ab \right)}{abc}=0\) \(ab+bc+ca=0\)

Summation Of Series In Trigonometry

1. \(\sin \alpha +\sin \left( \alpha +\beta \right)+\sin \left( \alpha +2\beta \right)+……….+\sin \left( \alpha +\left( n-1 \right)\beta \right)=\frac{\sin \left[ \alpha +\frac{\left( n-1 \right)}{2}\beta \right]\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}\)

When the angles of sine are in A.P. the sum of sine series is given by above formula.

However if \(\alpha =\beta\) in above case.

Then \(\sin \alpha +\sin 2\alpha +\sin 3\alpha +……….+\sin \alpha =\frac{\sin \left( \frac{n+1}{2} \right)\alpha \sin \frac{n\alpha }{2}}{\sin \frac{\alpha }{2}}\)

2. When cosine angles are in AP. \(AP.\,\,cos\alpha +cos2\alpha +\cos \alpha +……….+\cos n\alpha =\frac{\cos \left( \frac{n+1}{2} \right)\alpha \sin \frac{n\alpha }{2}}{\sin \frac{\alpha }{2}}\)

Remember:

If A,B,C are angles of \(\Delta\)

  1. \(\sin \left( B+C \right)=\sin A\)
  2. \(\cos \left( B+C \right)=-\cos A\)
  3. \(\sin \left( \frac{B+C}{2} \right)=\cos \frac{A}{2}\)
  4. \(\cos \left( \frac{B+C}{2} \right)=\sin \frac{A}{2}\)
  5. \(\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C\)
  6. \(\cos 2A+\cos 2B+\cos 2C=1-4\cos A\cos B\cos C\)
  7. \(\tan A+\tan B+\tan C=\tan A\tan B\tan C\)

Trignometric Equations

A solution of trigonometric equation is the value of the unknown angle that satisfy the equation.

Consider \(\sin \theta =\frac{1}{2}\) \(\theta =\frac{\pi }{6},\frac{5\pi }{6}\) etc.

Following are general solutions of trigonometric equation in the standard form:

Equation General Solution
\(\sin \theta =0 \) \(\theta =n\pi n\in z\)
\(\cos \theta =0\) \(\theta =\left( 2n+1 \right)\frac{\pi }{2}n\in z\)
\(\tan \theta =0\) \(\theta =n\pi ,n\in z\)
\(\sin \theta =\sin \alpha \) \(\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha ,n\in z\)
\(\cos \theta =\cos \alpha \) \(\theta =2n\pi \pm \alpha ,n\in z\)
\(\tan \theta =\tan \alpha\) \(\theta =n\pi +\alpha ,n\in z\)
\(\left. \begin{matrix} {{\sin }^{2}}\theta ={{\sin }^{2}}\alpha \\ {{\cos }^{2}}\theta ={{\cos }^{2}}\alpha \\ {{\tan }^{2}}\theta ={{\tan }^{2}}\alpha \\ \end{matrix} \right\} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \theta =n\pi \pm \alpha ,n\in z\)

The equation \(a\cos \theta +b\sin \theta =c\) is solvable for \(\left| c \right|\sqrt{{{a}^{2}}+{{b}^{2}}}\)

Example:

Solve \(4\sin x\sin 2x\sin 4x=\sin 3x\)

Solution:

\(4\sin x\sin \left( 3x-x \right)\sin \left( 3x+x \right)=\sin 3x\) \(4\sin x\left( {{\sin }^{2}}3x-{{\sin }^{2}}x \right)=\sin 3x\) \(4\sin x.{{\sin }^{2}}3x-4{{\sin }^{3}}x=3\sin x-4{{\sin }^{3}}x\) \(4\sin x{{\sin }^{2}}3x-3\sin x=0\) \(\sin x\left( 4{{\sin }^{2}}3x-3 \right)=0\) \(\sin x=0\;\; and \;\;{{\sin }^{2}}3x=\frac{3}{4}\) \(n\in z\) \({{\sin }^{2}}3x=\frac{3}{4}\) \({{\sin }^{2}}3x={{\left( \frac{\sqrt{3}}{4} \right)}^{2}}={{\sin }^{2}}\frac{\pi }{3}\) \(3x=m\pi \pm \frac{\pi }{3}m\in z\) \(x=\frac{m\pi }{3}\pm \frac{\pi }{9}\)

Example:

Solve \(\sin 3\theta +\cos 2\theta =0\) \(\cos 2\theta =-\sin 3\theta\) \(\cos 2\theta =-\sin 3\theta\)

Taking +ve sign

\(\theta =2n\pi +\frac{\pi }{2}+3\theta\) \(-\theta =2n\pi +\frac{\pi }{2}\) \(\theta =-2n\pi -\frac{\pi }{2}\)or

Taking -ve sign

\(\theta =2n\pi +\frac{\pi }{2}-3\theta n\in z\) \(4\theta =2n\pi +\frac{\pi }{2}\) \(\theta =\frac{n\pi }{2}+\frac{\pi }{8}\)

Example:

Solve \(\sqrt{3}\sec 2\theta =2\)

Solution:

\(\frac{\sqrt{3}}{\cos 2\theta }=2\) \(\cos 2\theta =\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}\) \(2\theta =2n\pi \pm \frac{\pi }{6}n\in z\) \(\theta =n\pi \pm \frac{\pi }{12}n\in z\)

Properties And Solutions Of Triangle

Solutions Of Triangle:

The side opposite to angle is denoted by opposite be is by and opposite to angle by

Sine Rule:

\(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\)

Circum Radius of the \(\Delta .\)

Cosine Rule:

\(\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\) \(\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\) \(\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\)

Area of \(\Delta =\frac{1}{2}bc\sin A\) \(=\frac{1}{2}ca\sin B\) \(=\frac{1}{2}ab\sin C\) \(=\sqrt{S\left( S-a \right)\left( S-b \right)\left( S-c \right)}\) \(=\frac{1}{2}\times\) base height

Projection Formula:

a=b cos c+c cos B

b=c cos A+a cos C

c=a cos B+b cos A

Napier’s Analogy:

This is also called as (Law of Tangents)

\(\tan \left( \frac{B-C}{3} \right)=\left( \frac{b-c}{b+c} \right)\cot \frac{A}{2}\) \(\tan \left( \frac{C-A}{2} \right)=\left( \frac{c-a}{c+a} \right)\cot \frac{B}{2}\) \(\tan \left( \frac{A-B}{2} \right)=\left( \frac{a-b}{a+b} \right)\cot \frac{C}{2}\)

Example:

If in a \(\Delta ABC\frac{a+b}{13}=\frac{a+c}{12}=\frac{b+c}{11}\)

Then \(P.T\;\;\;\;\frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}\)

Solution:

\(\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k\) \(b+c=11k,c+a=12k,a+b=13k\) \(2\left( a+b+c \right)=36k\) \(a+b+c=18k\)

b+c=11k

a=7k Similarly b=6k

c=5k

\(\cos A=\frac{{{b}^{2}}{{c}^{2}}-{{a}^{2}}}{2bc}=\frac{36{{k}^{2}}+25{{k}^{2}}-49{{k}^{2}}}{60{{k}_{2}}}=\frac{1}{5}\)

Similarly

\(\cos B=\frac{19}{35}\; and \; \cos C=\frac{5}{7}\) \(\cos A:\cos B:\cos C=\frac{1}{5}:\frac{19}{35}:\frac{8}{7}\)

=7:19:25

Circum Circle Of A Triangle

The circle passing through the vertices of a triangle is called circum circle of a triangle.

The radius is circum radius.

\(R=\frac{a}{2{sinA}}=\frac{b}{2\sin B}=\frac{c}{2\sin C}\)

Rso,

\(R=\frac{abc}{4\Delta }\) \(\Delta is \;the\; area\; of\; \Delta ABC\)

Inradius Of A Triangle:

\(r=\frac{\Delta }{s} \;where \;\Delta =\;area \;of \;\Delta\) \(s=\frac{a+b+c}{2}\) \(r=\left( s-a \right)\tan \frac{A}{2}=\left( s-b \right)\tan \frac{B}{2}=\left( s-c \right)\tan \frac{C}{2}\) \(r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\)

Example:

In a triangle \(ac2\sin \left( \frac{1}{2}A=C-B \right)=?\)

Solution:

We know that

A+B+C=180

A0-B+C=180-2B

\(2ac\sin \left( \frac{180-2B}{2} \right)=2ac\sin \left( 90-B \right)\)

=2ac cos B

\(={2ac}\left( \frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{{2ac}} \right)\) \(={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\)

Example:

In a \(\Delta \,\,ABC<<=60{}^\circ\)

Then prove that \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\)

Solution:

\(\cos c=\frac{1}{2}=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ba}\) \(ab={{a}^{2}}+{{b}^{2}}-{{c}^{2}}\rightarrow{{}}\left( i \right)\) \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\) \(\frac{a+b+c+c}{\left( a+c \right)\left( b+c \right)}=\frac{3}{a+b+c}\) \(\left( a+b+2c \right)\left( a+b+c \right)=3\left( a+c \right)\left( b+c \right)\) \({{a}^{2}}+{{b}^{2}}+2ab+2{{c}^{2}}+3ac+3bc=3ab+3ac+3bc+3{{c}^{2}}\) \({{a}^{2}}{{b}^{2}}-ab={{c}^{2}}\rightarrow{{}}\left( ii \right)\)

from (i) and (ii) we can say

\(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\)

Practise This Question

If 0<θ<90sec2θ+cosec2θ =