# 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

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

## 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:

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

### 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:

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

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

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.

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

Rso,

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

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

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

$\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$