Trigonometry Ratios

Trigonometry is one of the important branches of Mathematics. It is the study of triangles and its measurement. In this article, we come across the trigonometric ratios, graphs of trigonometric functions, identities, maximum and minimum values, main formulas and many more. 

Trigonometric Ratios

 

Trigonometric ratios

\(\begin{array}{l}\sin \theta =\frac{AB}{AC}=\frac{opp}{hyp}\end{array} \)

\(\begin{array}{l}\cos \theta =\frac{BC}{AC}=\frac{adj}{hyp}\end{array} \)

\(\begin{array}{l}\tan \theta =\frac{AB}{BC}=\frac{opp}{adj}\end{array} \)

\(\begin{array}{l}\cot\theta =\frac{BC}{AB}\end{array} \)

\(\begin{array}{l}\sec\theta =\frac{AC}{BC}\end{array} \)

\(\begin{array}{l}\cos ec\theta =\frac{AC}{AB}\end{array} \)

Trigonometric Circular Function

Trigonometric circular function

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

Cos

\(\begin{array}{l}\theta\end{array} \)
= x

Sin

\(\begin{array}{l}\theta\end{array} \)
= y

\(\begin{array}{l}\tan\theta=\frac{x}{y}\end{array} \)

Graphs of T Ratios

  • Sine

y = sin x

Graph of sine function

Domain R

Range (-1,1)

  • Cosine

y = cos x

Graph of cosine function

  • Tangent

y = tan x

Graph of tangent function

  • Co – tangent

y = cot x

Graph of co tangent function

 

  • Secant

\(\begin{array}{l}y=\sec x,\end{array} \)

Graph of secant function

 

\(\begin{array}{l}Domain:\,\,\,\,\,R-\left\{ (2x+1)\,\frac{\pi }{2} \right\}\end{array} \)
and

\(\begin{array}{l}Range:\,(-\infty -1)\, \cup \,(1,\infty )\end{array} \)

  • Cosecant

y = cosec x

Graph of co secant function

 

\(\begin{array}{l}Domain:\,R-\{n\pi \}\end{array} \)
and

\(\begin{array}{l}Range:\,(-\infty -1)\, \cup \,(1,\infty )\end{array} \)

Also Read:

Trigonometric Equations and its Solutions

Trigonometry JEE Main Previous Year Questions

Inverse Trig JEE Main Previous Year Questions

Trignometric Identities

\(\begin{array}{l}{{\sin }^{2 }\theta}+{{\cos }^{2 }\theta}=1\end{array} \)
.

\(\begin{array}{l}1+{{\tan }^{2 }\theta}=\,{{\sec }^{2 }\theta}\end{array} \)
.

\(\begin{array}{l}1+{{\cot }^{2 }\theta}=\cos e{{c}^{2 }\theta}\end{array} \)
.

\(\begin{array}{l}\sin ^{4}\theta +\cos^{4}=1-2\sin ^{2 }\theta\cos ^{2 }\theta\end{array} \)
.

\(\begin{array}{l}{{\sin }^{6}}\theta +{{\cos }^{6}}=1-3{{\sin }^{2 }\theta}{{\cos }^{2 }\theta}\end{array} \)

Remember:

\(\begin{array}{l}\sec \theta -\tan \theta =\frac{1}{\sec \theta +\tan \theta }\end{array} \)
and

\(\begin{array}{l}\cosec\theta -\cot \theta =\frac{1}{\cosec\theta +\cot \theta }\end{array} \)

T-Ratio At Some Standard Angles:

T ratio of standard angles

 

T-Ratio At Some Specific Angles:

T-ratio

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

Maximum and Minimum Value Of Standard Trigonometric Expressions

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

Problems on Trigonometry

Example 1: If

\(\begin{array}{l}{{\sec }^{4}}x-{{{cosec}}^{4}}x-2{{\sec }^{2}}x+2\cos e{{c}^{2}}x=\frac{15}{4}\end{array} \)
. Find tan x.

Solution:

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

or

\(\begin{array}{l}\left( {{\sec }^{4}}x-2{{\sec }^{2}}x+1 \right)-\left( \cos e{{c}^{4}}x-2\cos e{{c}^{2}}x+1 \right)=\frac{15}{4}\end{array} \)
,

\(\begin{array}{l}{{\left( {{\sec }^{2}}x-1 \right)}^{2}}-{{\left( \cos e{{c}^{2}}x-1 \right)}^{2}}=\frac{15}{4}\end{array} \)
,

\(\begin{array}{l}{{\left( {{\tan }^{2}}x \right)}^{2}}-{{\left( {{\cot }^{2}}x \right)}^{2}}=\frac{15}{4}\end{array} \)
,

\(\begin{array}{l}{{\tan }^{4}}x-\frac{1}{{{\tan }^{4}}x}=\frac{15}{4}\end{array} \)
,

\(\begin{array}{l}{{\tan }^{4}}x-\frac{1}{{{tan }^{4}}x}=4-\frac{1}{4}\end{array} \)

On comparing,

\(\begin{array}{l}{{\tan }^{4}}x=4\end{array} \)
,

\(\begin{array}{l}{{\tan }^{2}}x=2\end{array} \)
,

\(\begin{array}{l}tan x=\pm \sqrt{2}\end{array} \)

Example 2:

If

\(\begin{array}{l}\sin \theta ,\cos \theta ,\tan \theta\end{array} \)
are in G.P.

Then

\(\begin{array}{l}{{\cos }^{9}}\theta +{{\cos }^{6}}\theta +3{{\cos }^{3}}\theta {{\cos }^{2}}\theta -1=?\end{array} \)

Solution:

\(\begin{array}{l}\sin \theta .tan\theta =co{{s}^{2}}\theta\end{array} \)
,

\(\begin{array}{l}{{\sin }^{2}}\theta ={{\cos }^{3}}\theta\end{array} \)

Now,

\(\begin{array}{l}{{\left( {{\cos }^{3}}\theta \right)}^{3}}+{{\left( {{\cos }^{3}}\theta \right)}^{2}}+3{{\cos }^{3}}\theta {{\cos }^{2}}\theta -1\end{array} \)

=

\(\begin{array}{l}{{\sin }^{6}}\theta +{{\sin }^{4}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta -1\end{array} \)

=

\(\begin{array}{l}{{\sin }^{6}}\theta +{{\sin }^{4}}\theta -\left( 1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)\end{array} \)

=

\(\begin{array}{l}{{\sin }^{6}}\theta +{{\sin }^{4}}\theta -\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)\end{array} \)

=

\(\begin{array}{l}{{\sin }^{4}}\theta -{{\cos }^{6}}\theta\end{array} \)

=

\(\begin{array}{l}{{\sin }^{4}}\theta -{{\left( {{\cos }^{3}}\theta \right)}^{2}}\end{array} \)

=

\(\begin{array}{l}{{\sin }^{4}}\theta -{{\sin }^{4}}\theta =0\end{array} \)

Example 3:

If

\(\begin{array}{l}\sec x-\tan x=P\end{array} \)
then sec x = ?

Solution:

Given:

\(\begin{array}{l}\sec x\;-\;tan x=P\end{array} \)
….(1)

We know,

\(\begin{array}{l}\sec x-\tan x=\frac{1}{\sec x+\tan x}\end{array} \)

or

\(\begin{array}{l}\sec x\;+\;tan x=\frac{1}{P}\end{array} \)
….(2)

Adding (1) and (2)

\(\begin{array}{l}{2\sec x=P+\frac{1}{P}}\end{array} \)

or

\(\begin{array}{l}\sec x=\frac{{{P}^{2}}+1}{2P}\end{array} \)

Example 4:

Prove that

\(\begin{array}{l}\sin \left( -420 \right).\cos \left( 390 \right)+\cos \left( -660 \right)\sin \left( 330 \right)=-1\end{array} \)

Solution:

\(\begin{array}{l}-\sin \left( 420 \right)\cos \left( 390 \right)+\cos \left( 660 \right)\sin 330\end{array} \)

=

\(\begin{array}{l}-\sin \left( 2\pi +60 \right)\cos \left( 2\pi +30 \right)+\cos \left( 4\pi -60 \right)\sin \left( 2\pi -30 \right)\end{array} \)

=

\(\begin{array}{l}-\sin 60\cos 30+\cos 60\left( -\sin 30 \right)\end{array} \)

=

\(\begin{array}{l}-\frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}-\frac{1}{2}.\frac{1}{2}\end{array} \)

=

\(\begin{array}{l}-\frac{3}{4}-\frac{1}{4}=-\frac{4}{4}=-1.\end{array} \)

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

Remember:

If A+B+C = 0 then

\(\begin{array}{l}\tan A+\tan B+\tan C=\tan A\tan B\tan C\end{array} \)

Let us solve some of the problems to understand the concept in a better way.

Example 1:

If

\(\begin{array}{l}\sin \alpha =\frac{3}{5}\And \cos \beta =\frac{9}{41}\alpha ,\beta \in\end{array} \)
I quadrant.

Find

\(\begin{array}{l}\sin \left( \alpha -\beta \right)\end{array} \)

Solution:

\(\begin{array}{l}\sin \alpha =\frac{3}{5}\end{array} \)
,

\(\begin{array}{l}{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\end{array} \)
,

\(\begin{array}{l}{{\cos }^{2}}\alpha =1-\frac{9}{25}=\frac{16}{25}\end{array} \)
,

\(\begin{array}{l}\cos \alpha =\frac{4}{5},\frac{-4}{5}\end{array} \)
as in first quadrant.

\(\begin{array}{l}\cos \beta =\frac{9}{41}\end{array} \)
,

\(\begin{array}{l}{{\cos }^{2}}\beta +{{\sin }^{2}}\beta =1\end{array} \)
,

\(\begin{array}{l}{{\sin }^{2}}\beta =1-\frac{81}{1681}=\frac{1600}{1681}\end{array} \)
,

\(\begin{array}{l}\sin \beta =\frac{40}{41}\end{array} \)
,

\(\begin{array}{l}\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta\end{array} \)
,

\(\begin{array}{l}=\frac{3}{5}\times \frac{9}{41}-\frac{4}{5}\times \frac{40}{41}=\frac{27-160}{205}\end{array} \)
,

\(\begin{array}{l}=\frac{-133}{205}.\end{array} \)

Example 2:

Find value of

\(\begin{array}{l}\frac{\sin \left( B-C \right)}{\cos B\cos C}+\frac{\sin \left( C-A \right)}{\cos C\cos A}+\frac{\sin \left( A-B \right)}{\cos A\cos B}=?\end{array} \)

Solution:

\(\begin{array}{l}\frac{\sin B\cos C-\cos B\sin C}{\cos B\cos C}+\frac{\sin C\cos A-\cos C\sin A}{\cos C\cos A}+\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B}\end{array} \)
,

\(\begin{array}{l}tan \;B\;-\;tan \;C\;+\;tan C\;-\;tan A\;+\;tan A\;-\;tan\; B = 0\end{array} \)

Transformation Formulae:

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

T-Ratio Of Multiple Angles

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

Remember:

\(\begin{array}{l}1+\cos 2A=2{{\cos }^{2}}A\end{array} \)

 

\(\begin{array}{l}1-\cos 2A=2{{\sin }^{2}}A\end{array} \)

 

\(\begin{array}{l}\frac{1-\cos 2A}{1+\cos 2A}={{\tan }^{2}}A\end{array} \)

Values Of T-Ratios At Some Useful Angles

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

Important Results:

\(\begin{array}{l}\sin \theta \sin \left( 60-\theta \right)\sin \left( 60+\theta \right)=\frac{1}{4}\sin 3\theta\end{array} \)
,

\(\begin{array}{l}\cos \theta \cos \left( 60-\theta \right)\cos \left( 60+\theta \right)=\frac{1}{4}\cos3 \theta\end{array} \)
,

\(\begin{array}{l}\tan \theta \tan \left( 60-\theta \right)\tan \left( 60+\theta \right)=\tan 3\theta\end{array} \)

Lets solve few more examples:

Example 1: Find value of

\(\begin{array}{l}\frac{\sin A\sin 2A+\sin 3A\sin 6A+\sin 4A\sin 13A}{2\sin A\cos 2A+sin3Acos6A+sin4Acos13A}=?\end{array} \)

Solution:

\(\begin{array}{l}\frac{\cos \left( -A \right)-{\cos \left( 3A \right)}+{\cos 3A}-{\cos 9A}+{\cos 9A}-\cos 17A}{{\sin 3A}+\sin \left( -A \right)+{\sin 9A}+{sin\left( -3A \right)}+\sin \left( 17A \right)+{\sin \left( -9A \right)}}\end{array} \)

=

\(\begin{array}{l}\frac{\cos A-\cos 17A}{\sin 17A-\sin A}\end{array} \)

=

\(\begin{array}{l}\frac{{2}\sin 9A{\sin 8A}}{{2}\cos 9A{\sin 8A}}\end{array} \)

=tan 9A

Example 2:

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

Solution:

\(\begin{array}{l}{{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +……….+{{\sin }^{2}}85{}^\circ +{{\sin }^{2}}90{}^\circ\end{array} \)

=

\(\begin{array}{l}{{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +……….+{{\cos }^{2}}5{}^\circ +{{\sin }^{2}}90{}^\circ\end{array} \)

=

\(\begin{array}{l}\left( {{\sin }^{2}}5{}^\circ +{{\cos }^{2}}5{}^\circ \right)+………. +{{\sin }^{2}}45{}^\circ+{{\sin }^{2}}90{}^\circ\end{array} \)

=[1+1+…+1] + (1/√2)2 + 1

=

\(\begin{array}{l}8\times 1+1+\frac{1}{2}\end{array} \)

=

\(\begin{array}{l}8+1+\frac{1}{2}=9\frac{1}{2}\end{array} \)

Example 3:

Prove that

\(\begin{array}{l}\sin \alpha +sin\left( \alpha +\frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{4\pi }{3} \right)=0\end{array} \)

Solution:

\(\begin{array}{l}2\sin \left( \frac{2\alpha }{2}+\frac{4\pi }{6} \right)\cos \left( -\frac{4\pi }{6} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right)\end{array} \)

\(\begin{array}{l}2\sin \left( \alpha +\frac{2\pi }{3} \right)\cos \left( \frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right)\end{array} \)

\(\begin{array}{l}2\times \left( -\frac{1}{2} \right)\sin \left( \alpha +\frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right)\end{array} \)

\(\begin{array}{l}-\sin \left( \alpha +\frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right)\end{array} \)

= 0

Example 4:

If

\(\begin{array}{l}a\sin \theta =b\sin \left( \theta +\frac{2\pi }{3} \right)=c\sin \left( \theta +\frac{4\pi }{3} \right)\end{array} \)
.

Prove that ab + bc + ca = 0.

Solution:

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

\(\begin{array}{l}\sin \theta =\frac{k}{a}\end{array} \)
,

\(\begin{array}{l}\sin \left( \theta +\frac{2\pi }{3} \right)=\frac{k}{b}\end{array} \)
,

\(\begin{array}{l}\sin \left( \theta +\frac{4\pi }{3} \right)=\frac{k}{c}\end{array} \)

We know

\(\begin{array}{l}\sin \theta +\sin \left( \theta +\frac{2\pi }{3} \right)+\sin \left( \theta +\frac{4\pi }{3} \right)=0\end{array} \)
,

\(\begin{array}{l}\frac{k}{a}+\frac{k}{b}+\frac{k}{c}=0\end{array} \)
,

\(\begin{array}{l}\frac{k\left( bc+ac+ab \right)}{abc}=0\end{array} \)
,

\(\begin{array}{l}ab+bc+ca=0\end{array} \)

Summation Of Series In Trigonometry

1.

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

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

However if

\(\begin{array}{l}\alpha =\beta\end{array} \)
in above case.

Then

\(\begin{array}{l}\sin \alpha +\sin 2\alpha +\sin 3\alpha +……….+\sin \alpha =\frac{\sin \left( \frac{n+1}{2} \right)\alpha \sin \frac{n\alpha }{2}}{\sin \frac{\alpha }{2}}\end{array} \)

2. When cosine angles are in AP.

\(\begin{array}{l}cos\alpha +cos2\alpha +\cos \alpha +……….+\cos n\alpha =\frac{\cos \left( \frac{n+1}{2} \right)\alpha \sin \frac{n\alpha }{2}}{\sin \frac{\alpha }{2}}\end{array} \)

Remember:

If A,B,C are angles of

\(\begin{array}{l}\Delta\end{array} \)

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

Trigonometric Equations

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

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

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

\(\begin{array}{l}\left. \begin{matrix} {{\sin }^{2}}\theta ={{\sin }^{2}}\alpha \\ {{\cos }^{2}}\theta ={{\cos }^{2}}\alpha \\ {{\tan }^{2}}\theta ={{\tan }^{2}}\alpha \\ \end{matrix} \right\} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \theta =n\pi \pm \alpha ,n\in z\end{array} \)

Example 1:

Solve

\(\begin{array}{l}4\sin x\sin 2x\sin 4x=\sin 3x\end{array} \)

Solution:

\(\begin{array}{l}4\sin x\sin \left( 3x-x \right)\sin \left( 3x+x \right)=\sin 3x\end{array} \)
,

\(\begin{array}{l}4\sin x\left( {{\sin }^{2}}3x-{{\sin }^{2}}x \right)=\sin 3x\end{array} \)
,

\(\begin{array}{l}4\sin x.{{\sin }^{2}}3x-4{{\sin }^{3}}x=3\sin x-4{{\sin }^{3}}x\end{array} \)
,

\(\begin{array}{l}4\sin x{{\sin }^{2}}3x-3\sin x=0\end{array} \)
,

\(\begin{array}{l}\sin x\left( 4{{\sin }^{2}}3x-3 \right)=0\end{array} \)
,

\(\begin{array}{l}\sin x=0\;\; and \;\;{{\sin }^{2}}3x=\frac{3}{4}\end{array} \)
,

\(\begin{array}{l}n\in z\end{array} \)
,

\(\begin{array}{l}{{\sin }^{2}}3x=\frac{3}{4}\end{array} \)
,

\(\begin{array}{l}{{\sin }^{2}}3x={{\left( \frac{\sqrt{3}}{4} \right)}^{2}}={{\sin }^{2}}\frac{\pi }{3}\end{array} \)
,

\(\begin{array}{l}3x=m\pi \pm \frac{\pi }{3}m\in z\end{array} \)
,

\(\begin{array}{l}x=\frac{m\pi }{3}\pm \frac{\pi }{9}\end{array} \)

Example 2: Solve

\(\begin{array}{l}\sin 3\theta +\cos 2\theta =0\end{array} \)

Solution:

\(\begin{array}{l}\cos 2\theta =-\sin 3\theta\end{array} \)
.

Taking +ve sign

\(\begin{array}{l}\theta =2n\pi +\frac{\pi }{2}+3\theta\end{array} \)
,

\(\begin{array}{l}-\theta =2n\pi +\frac{\pi }{2}\end{array} \)
,

\(\begin{array}{l}\theta =-2n\pi -\frac{\pi }{2}\end{array} \)
or

Taking -ve sign

\(\begin{array}{l}\theta =2n\pi +\frac{\pi }{2}-3\theta; n\in z\end{array} \)
,

\(\begin{array}{l}4\theta =2n\pi +\frac{\pi }{2}\end{array} \)
,

\(\begin{array}{l}\theta =\frac{n\pi }{2}+\frac{\pi }{8}\end{array} \)

Example 3: Solve

\(\begin{array}{l}\sqrt{3}\sec 2\theta =2\end{array} \)

Solution:

\(\begin{array}{l}\frac{\sqrt{3}}{\cos 2\theta }=2\end{array} \)
,

\(\begin{array}{l}\cos 2\theta =\frac{\sqrt{3}}{2}=\cos \frac{\pi }{6}\end{array} \)
,

\(\begin{array}{l}2\theta =2n\pi \pm \frac{\pi }{6}n\in z\end{array} \)
,

\(\begin{array}{l}\theta =n\pi \pm \frac{\pi }{12}n\in z\end{array} \)

Properties And Solutions Of Triangle

Solutions Of Triangle:

Solutions of triangle

 

Here A, B and C are the angles and a, b and c are the lengths of sides of the triangle ABC.

Sine Rule:

Sine rule relates the length of sides of a triangle to the sines of its angles. 

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

where 2R is the diameter of the triangle’s circumcircle.

Cosine Rule:

\(\begin{array}{l}\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\end{array} \)
and

\(\begin{array}{l}\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\end{array} \)
and

\(\begin{array}{l}\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\end{array} \)

Area of

\(\begin{array}{l}\Delta =\frac{1}{2}bc\sin A\end{array} \)

\(\begin{array}{l}=\frac{1}{2}ca\sin B\end{array} \)

\(\begin{array}{l}=\frac{1}{2}ab\sin C\end{array} \)

\(\begin{array}{l}=\sqrt{S\left( S-a \right)\left( S-b \right)\left( S-c \right)}\end{array} \)

\(\begin{array}{l}=\frac{1}{2}\times base\times height\end{array} \)

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)

\(\begin{array}{l}\tan \left( \frac{B-C}{3} \right)=\left( \frac{b-c}{b+c} \right)\cot \frac{A}{2}\end{array} \)
,

\(\begin{array}{l}\tan \left( \frac{C-A}{2} \right)=\left( \frac{c-a}{c+a} \right)\cot \frac{B}{2}\end{array} \)
,

\(\begin{array}{l}\tan \left( \frac{A-B}{2} \right)=\left( \frac{a-b}{a+b} \right)\cot \frac{C}{2}\end{array} \)

Example:

In a triangle ABC,

\(\begin{array}{l}\frac{a+b}{13}=\frac{a+c}{12}=\frac{b+c}{11}\end{array} \)

Then prove that

\(\begin{array}{l}\frac{\cos A}{7}=\frac{\cos B}{19}=\frac{\cos C}{25}\end{array} \)

Solution:

\(\begin{array}{l}\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k\end{array} \)
,

\(\begin{array}{l}b+c=11k,c+a=12k,a+b=13k\end{array} \)
,

\(\begin{array}{l}2\left( a+b+c \right)=36k\end{array} \)
,

\(\begin{array}{l}a+b+c=18k\end{array} \)

b+c=11k

a=7k Similarly b=6k

c=5k

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

Similarly

\(\begin{array}{l}\cos B=\frac{19}{35}\; and \; \cos C=\frac{5}{7}\end{array} \)
,

\(\begin{array}{l}\cos A:\cos B:\cos C=\frac{1}{5}:\frac{19}{35}:\frac{8}{7}\end{array} \)

=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.

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

Also,

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

Inradius Of A Triangle:

\(\begin{array}{l}r=\frac{\Delta }{s} \;where \;\Delta =\;area \;of \;\Delta\end{array} \)
,

\(\begin{array}{l}s=\frac{a+b+c}{2}\end{array} \)
,

\(\begin{array}{l}r=\left( s-a \right)\tan \frac{A}{2}=\left( s-b \right)\tan \frac{B}{2}=\left( s-c \right)\tan \frac{C}{2}\end{array} \)
,

\(\begin{array}{l}r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\end{array} \)

Example:

In a triangle ABC, angle C = 60 degrees.

Then prove that

\(\begin{array}{l}\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\end{array} \)

Solution:

\(\begin{array}{l}\cos c=\frac{1}{2}=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ba}\end{array} \)
,

\(\begin{array}{l}ab={{a}^{2}}+{{b}^{2}}-{{c}^{2}}\rightarrow{{}}\left( i \right)\end{array} \)
,

\(\begin{array}{l}\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\end{array} \)
,

\(\begin{array}{l}\frac{a+b+c+c}{\left( a+c \right)\left( b+c \right)}=\frac{3}{a+b+c}\end{array} \)
,

\(\begin{array}{l}\left( a+b+2c \right)\left( a+b+c \right)=3\left( a+c \right)\left( b+c \right)\end{array} \)
,

\(\begin{array}{l}{{a}^{2}}+{{b}^{2}}+2ab+2{{c}^{2}}+3ac+3bc=3ab+3ac+3bc+3{{c}^{2}}\end{array} \)
,

\(\begin{array}{l}{{a}^{2}}{{b}^{2}}-ab={{c}^{2}}\rightarrow{{}}\left( ii \right)\end{array} \)

from (i) and (ii) we can say

\(\begin{array}{l}\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\end{array} \)

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Frequently Asked Questions

What is trigonometry?

Trigonometry is a branch of mathematics that deals with the relationship between the sides of a triangle (right triangle) with its angles.

Name the six basic trigonometric functions.

The basic trigonometric functions are sin, cos, tan, sec, cosec, and cot.

Give the Pythagorean trigonometric identities.

sin2a + cos2a = 1
1 + tan2a = sec2a
1 + cot2 a = cosec2a

Give an application of trigonometry.

The height of an object or the distance between two distinct objects can be calculated using trigonometric ratios.

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