# Inverse Trigonometric Functions

Here students will get familiar with the topic of inverse trigonometric functions and all the related terms associated with it. They will also find solved problems based on this concept that will further help students understand the topic more clearly.

## Introduction to Inverse Trigonometric Functions

If $\sin \frac{\pi }{6}=\frac{1}{2}$ we can write that $\frac{\pi }{6}=\frac{1}{2}={{\sin }^{-1}}\left( \frac{1}{2} \right)$ hence ${{\cos }^{-1}}x$means the angle whose cosine is x. Trigonometric functions are many-one hence for inverse functions to be defined. We restrict the domain of trigonometric functions, so as to make them one.

For ${{\sin }^{-1}}x$ we choose $\left[ \frac{-\pi }{2},\frac{\pi }{2} \right]$ as the principle value branch.

### Domain And Range Of Inverse Trigonometric Functions With Graphs

1. $f(x)=si{{n}^{-1}}x$

$D\in [-1,1]$

$R:\left[ \frac{-\pi }{2},\frac{\pi }{2} \right]$

2. $f(x)={{\cos }^{-1}}x$

$D\in [-1,1]$

$R:\left[ 0,\pi \right]$

3. $f(x)={{\tan }^{-1}}x$

$D\in R$

$R\in \left[ \frac{-\pi }{2},\frac{\pi }{2} \right]$

4. $f(x)={{\cot }^{-1}}x$

$D\in R$

$R\in \left[ 0,\pi \right]$

5. $f(x)={{\sec }^{-1}}x$

$D\in (-\infty ,-1]\cup [1,\infty )$

$R\in \left[ 0,\pi \right]-\left\{ \frac{\pi }{2} \right\}$

$\\f(x)=\cos e{{c}^{-1}}x \\ D\in (-\infty ,-1]\cup [1,\infty ] \\ R\in \left[ \frac{-\pi }{2},\frac{\pi }{2} \right]-\left\{ 0 \right\} \\$.

### Solved Examples Of Inverse Trigonometric Functions

Example 1. Find principal value of ${{\sin }^{-1}}\left( \frac{-1}{2} \right)$

Sol:

Let ${{\sin }^{-1}}\left( \frac{-1}{2} \right)=y$

$\sin y=\frac{-1}{2}$

∴ $y=\frac{-\pi }{6}$

Example 2. $\tan \left( \frac{1}{2}{{\cos }^{-1}}\frac{\sqrt{5}}{3} \right)=?$

Sol:

Let $\tan \left( \frac{1}{2}{{\cos }^{-1}}\frac{\sqrt{5}}{3} \right)=\tan \theta$

∴ $\,\,\,\,2\theta ={{\cos }^{-1}}\frac{\sqrt{5}}{3}$

∴ $\,\,\,\,\cos \,2\theta =\frac{\sqrt{5}}{3}$

∴ $\tan \theta =\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}=\frac{3-\sqrt{5}}{2}\,Ans.$

Example 3: ${{\cot }^{-1}}(-1)=?$

Sol:

Say ${{\cot }^{-1}}(-1)=\alpha$

∴ $\\cot \alpha =-1$

$\alpha =\frac{3\pi }{4}$

### Properties Of Inverse Trigonometric Functions

Property 1: ${{\sin }^{-1}}(x)=cose{{c}^{-1}}\left( \frac{1}{x} \right)\,\,\,\,x\in [-1,1]-\{0\}$

Property 2: ${{\cos }^{-1}}(x)={{\sec }^{-1}}\left( \frac{1}{x} \right)\,\,\,\,x\in [-1,1]-\{0\}$

Property 3: ${{\tan }^{-1}}x=\left\{ \begin{matrix} {{\cot }^{-1}}\left( \frac{1}{x} \right)\,\,\,\,\,\,\,\,\,if\,x>0 \\ {{\cot }^{-1}}\left( \frac{1}{x} \right)-\pi \,\,\,\,\,if\,x<0\,\,\, \\ \end{matrix} \right.$

Property 4: ${{\cot }^{-1}}x=\left\{ \begin{matrix} {{\tan }^{-1}}\left( \frac{1}{x} \right)\,\,\,\,\,\,\,\,\,if\,x>0 \\ {{\tan }^{-1}}\left( \frac{1}{x} \right)+\pi \,\,\,\,\,if\,x<0\,\,\, \\ \end{matrix} \right.$

Property 5:

Illustration

• $sin^{-1}\left ( \frac{1}{3} \right ) = cosec^{-1}(3)$
• $cos^{-1}\left ( \frac{1}{4} \right ) = sec^{-1}(4)$
• $sin^{-1}\left ( \frac{-3}{4} \right )=cosec^{-1}\left ( \frac{-4}{3} \right )=sin^{-1}\left ( \frac{3}{4} \right )$
• $tan^{-1}(-3) = cot^{-1}\left ( -\frac{1}{3} \right )-\pi$

Property Set 6:

$\\(i) {{\sin }^{-1}}(cos\theta )=\frac{\pi }{2}-\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,\theta \in \,[0,\pi ]\\ (ii) {{\cos }^{-1}}(sin\theta )=\frac{\pi }{2}-\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,\theta \in \,\left[ \frac{-\pi }{2},\frac{\pi }{2} \right]\\ (iii) {{\tan }^{-1}}(cot\theta )=\frac{\pi }{2}-\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta \in \,\left[ 0,\pi \right]\\ (iv) {{\cot }^{-1}}(tan\theta )=\frac{\pi }{2}-\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta \in \,\left[ \frac{-\pi }{2},\frac{\pi }{2} \right]$

$\\(v) {{\sec }^{-1}}(cosec\theta )=\frac{\pi }{2}-\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta \in \,\left[ \frac{-\pi }{2},0 \right]\cup \left[ 0,\frac{\pi }{2} \right]\\ (vi) \cos e{{c}^{-1}}(sec\theta )=\frac{\pi }{2}-\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta \in \,\left[ 0,\pi \right]-\left\{ \frac{\pi }{2} \right\}\\ (vii) {{\sin }^{-1}}(x)={{\cos }^{-1}}\left( \sqrt{1-{{x}^{2}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,0\le x\le 1 =-{{\cos }^{-1}}\left( \sqrt{1-{{x}^{2}}} \right)\,\,\,\,\,\,\,\,\,\,-1\le x<0\\$

1. ${{\sin }^{-1}}\left( -\frac{1}{2} \right)={{\tan }^{-1}}\left( \frac{\frac{-1}{2}}{\sqrt{1\frac{-1}{4}}} \right)\,\,\,\,\,\,\,\,\,\,by\,u\;sin g\;,triangle\,as\,below$

∴ $\,\,\,\,\,{{\sin }^{-1}}\left( \frac{-1}{2} \right)={{\tan }^{-1}}\left( \frac{-1}{\sqrt{3}} \right)$

1. ${{\cos }^{-1}}\left( \frac{-3}{4} \right)=\pi -{{\sin }^{-1}}A\,\,\,\,\,Find\,A\,$

Now,

∴ $\,\,\,\,\,\,=\pi -{{\sin }^{-1}}\frac{\sqrt{7}}{4}\,Ans$

1. ${{\cos }^{-1}}\left( \frac{1}{4} \right)={{\sin }^{-1}}\sqrt{1-\frac{1}{16}}=\,{{\sin }^{-1}}\left( \frac{\sqrt{15}}{4} \right)$.
2. ${{\sin }^{-1}}\left( -\frac{1}{2} \right)=-{{\cos }^{-1}}\sqrt{1-\frac{1}{4}}=\,-{{\cos }^{-1}}\frac{\sqrt{3}}{2}$
3. ${{\sin }^{2}}\left( {{\tan }^{-1}}\left( \frac{3}{4} \right) \right)$

$={{\sin }^{2}}\left( si{{n}^{-1}}\left( \frac{3}{5} \right) \right)$

$={{\left( \frac{3}{5} \right)}^{2}}=\frac{9}{25}Ans.$

1. ${{\sin }^{-1}}\left( \sin \frac{2\pi }{3} \right)=\frac{\pi }{3}$
2. ${{\cos }^{-1}}\left( \cos \frac{4\pi }{3} \right)=\frac{2\pi }{3}$
3. ${{\sin }^{-1}}\left( \cos \frac{33\pi }{10} \right)={{\sin }^{-1}}\cos \left( 3\pi +\frac{3\pi }{10} \right)$

$={{\sin }^{-1}}\left( -\sin \left( \frac{\pi }{2}-\frac{3\pi }{10} \right) \right)$

$=-\left( \frac{\pi }{2}-\frac{3\pi }{10} \right)=\frac{-\pi }{5}$

Property set 5

(i) ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2}$

(ii) ${{\tan }^{-1}}x+{{\cot }^{-1}}(x)=\frac{\pi }{2}$

(iii) ${{\sec }^{-1}}x+{{{cosec}}^{-1}}x=\frac{\pi }{2}$

(iv) ${{\tan }^{-1}}x+{{\tan }^{-1}}\left( \frac{1}{x} \right)=\frac{\pi }{2}x>0-\frac{\pi }{2}x<0$

e.g. ${{\sec }^{-1}}(4)+cose{{c}^{-1}}(4)=\frac{\pi }{2}$

${{\tan }^{-1}}(3)+co{{t}^{-1}}(3)=\frac{\pi }{2}$

Property set 6

(1) If x, y > 0

${{\tan }^{-1}}x+{{\tan }^{-1}}y=\left\{ \begin{matrix} {{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)xy<1 \\ \pi +{{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)xy>1 \\ \end{matrix} \right.$

(2) if x, y < 0

${{\tan }^{-1}}x+{{\tan }^{-1}}y=\left\{ \begin{matrix} {{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)xy<1 \\ -\pi +{{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)xy<1 \\ \end{matrix} \right.$

(3) ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right)xy>$

$=\,\pi +{{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right)\begin{matrix} x>0 \\ y<0 \\ \end{matrix}$

$=-\pi +{{\tan }^{-1}}\left( \frac{x-y}{1+xy} \right)\begin{matrix} x<0 \\ y>0 \\ \end{matrix}$

Illustration

1. ${{\tan }^{-1}}\left( \frac{-1}{2} \right)+{{\tan }^{-1}}\left( -\frac{1}{3} \right)={{\tan }^{-1}}\frac{\left( -\frac{1}{2}-\frac{1}{3} \right)}{1-\frac{1}{6}}$

$={{\tan }^{-1}}(-1)$

$=-\frac{\pi }{4}\,Ans.$

2. ${{\tan }^{-1}}(-2)+{{\tan }^{-1}}(-3)={{\tan }^{-1}}\frac{-2+-3}{1-6}$

$={{\tan }^{-1}}\frac{-5}{-5}={{\tan }^{-1}}1,-\pi$

= $=-\frac{3\pi }{4}\,Ans.$

3. ${{\tan }^{-1}}(-3)+{{\tan }^{-1}}\left( -\frac{1}{3} \right)=-(ta{{n}^{-1}}B)+ta{{n}^{-1}}\left( \frac{1}{3} \right)$

$=-\frac{\pi }{2}$

4. ${{\tan }^{-1}}\left( \frac{5}{3} \right)-ta{{n}^{-1}}\left( \frac{1}{4} \right)=\tan -1\left( \frac{\frac{5}{3}-\frac{1}{4}}{1+\frac{5}{12}} \right)$

=${{\tan }^{-1}}\frac{17}{17}$

$={{\tan }^{-1}}1=\frac{\pi }{4}$

5. ${{\tan }^{-1}}2x+{{\tan }^{-1}}3x=\frac{\pi }{4}$

⇒ $\,\,{{\tan }^{-1}}\left( \frac{5x}{1-6{{x}^{2}}} \right)=\frac{\pi }{4}$

⇒ $\,\,\,\frac{5x}{1-6{{x}^{2}}}=1$

⇒ $\,\,\,6{{x}^{2}}-5x+1=0\,\,\,\Rightarrow \,\,\,x=\frac{1}{6}or-1$

∴ $\,\,\,x=\frac{1}{6}\,\,\,\,\,\,\,\,\,\,\,\,\,x=-1\,rejected$

6. If ${{\tan }^{-1}}(4)+ta{{n}^{-1}}(5)=co{{t}^{-1}}(\lambda ).Find\,\lambda$

${{\tan }^{-1}}\left( \frac{9}{1-20} \right)={{\cot }^{-1}}\lambda$

${{\tan }^{-1}}\left( \frac{9}{19} \right)={{\cot }^{-1}}(\lambda )$

$-{{\tan }^{-1}}\left( \frac{9}{19} \right)={{\cot }^{-1}}(\lambda )$

$-{{\cot }^{-1}}\frac{9}{19}={{\cot }^{-1}}(\lambda )$

$\lambda =\frac{-19}{9}$

Property set 8

${{\sin }^{-1}}(x)+si{{n}^{-1}}(y)=si{{n}^{-1}}\left( x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right)for\,{{x}^{2}}+{{y}^{2}}\le 1$

${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right)$

1. ${{\sin }^{-1}}\left( \frac{4}{5} \right)+{{\sin }^{-1}}\left( \frac{7}{25} \right)={{\sin }^{-1}}(A)find\,A$

$={{\sin }^{-1}}\left( \frac{4}{5}\sqrt{1-{{\left( \frac{7}{25} \right)}^{2}}}+\sqrt{^{1-}{{\left( \frac{4}{5} \right)}^{2}}}\frac{7}{25} \right)$

$={{\sin }^{-1}}\left( \frac{117}{125} \right)$

2. Prove that ${{\sin }^{-1}}\left( \frac{4}{5} \right)+\sin \left( \frac{5}{13} \right)+{{\sin }^{-1}}\left( \frac{16}{65} \right)=\frac{\pi }{2}$

Ans: ${{\sin }^{-1}}\left( \frac{63}{65} \right)+{{\sin }^{-1}}\left( \frac{16}{65} \right)$

= ${{\cos }^{-1}}\left( \frac{16}{65} \right)+{{\sin }^{-1}}\left( \frac{16}{65} \right)$

$=\frac{\pi }{2}$

Property set 9 Corresponding graphs

1. ${{\sin }^{-1}}(sin\,x)=-\pi -\pi \,\,\,\,\,\,\,\,\,\,\,if\,x\in \left[ -\frac{3\pi }{2},\frac{-\pi }{2} \right]$

$=\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,x\in \left[ \frac{-\pi }{2},\frac{\pi }{2} \right]$

$=\,\pi -x\,\,\,\,\,\,\,\,\,\,\,\,\,if\,\,x\in \left[ \frac{\pi }{2},\frac{3\pi }{2} \right]$

$=\,-2\pi +x\,\,\,\,\,if\,\,\,x\in \,\left[ \frac{3\pi }{2},\frac{5\pi }{2} \right]$

And so on

2. ${{\cos }^{-1}}(cosx)=2\pi +x\,\,\,\,\,\,if\,x\in [-2\pi ,-\pi ]$

$=-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\in [-\pi ,0]$

$=x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\in \,[0,\pi ]$

$=2\pi -x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\in [\pi ,2\pi ]$

$=-2\pi +x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\in [2\pi ,3\pi ]$

3. ${{\tan }^{-1}}\tan x=\pi +x\,\,\,\,\,\,\,x\in \left( \frac{-3\pi }{2},\frac{-\pi }{2} \right)$

$x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \frac{-\pi }{2},\frac{\pi }{2} \right)$

$x-\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \frac{\pi }{2},\frac{3\pi }{2} \right)$

$x-2\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \frac{3\pi }{2},\frac{5\pi }{2} \right)$

4. ${{\sin }^{-1}}\left( \sin \frac{2\pi }{3} \right)=\pi -\frac{2\pi }{3}=\frac{\pi }{3}$

5. ${{\cos }^{-1}}\left( \cos \left( \frac{13\pi }{6} \right) \right)={}^{\pi }/{}_{6}$

6. ${{\sin }^{-1}}\sin (4)=\pi -4$

7. ${{\sin }^{-1}}\sin (6)=6-2\pi$

$\\8. {{\sin }^{-1}}\sin (12)=12-4\pi \\ Q. {{\cos }^{-1}}(cos3)=3\\ Q. {{\cos }^{-1}}(cos5)=2\pi -5\\ Q. {{\cos }^{-1}}(cos6)=2\pi -6\\ Q. {{\tan }^{-1}}(tan3)=3-\pi$

Property set 10

1. $2{{\sin }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)$

2. $2{{\cos }^{-1}}x={{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)$

3. $2{{\tan }^{-1}}x={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)$

4. ${{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x$

5. $3{{\sin }^{-1}}x={{\sin }^{-1}}(3x-4{{x}^{3}})$

6. $3{{\cos }^{-1}}x={{\cos }^{-1}}\left( 4{{x}^{3}}-3x \right)$

7. $3{{\tan }^{-1}}x={{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)$

8. $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)$

9. ${{\tan }^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right)=\frac{1}{2}{{\tan }^{-1}}x$

1. f(x) = ${{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)+2{{\tan }^{-1}}x.$

If x > 1 find cos (f(10))

Ans. $f(10)=si{{n}^{-1}}\left( \frac{20}{101} \right)+2{{\tan }^{-1}}(10)$

$={{\tan }^{-1}}\left( \frac{20}{99} \right)+2{{\tan }^{-1}}(10)$

$=\pi +{{\tan }^{-1}}\left( \frac{20}{99} \right)\pm {{\tan }^{-1}}\left( \frac{20}{99} \right)$

= π

2. Find $\tan \left( {{\cos }^{-1}}\left( \frac{4}{5} \right)+{{\tan }^{-1}}\left( \frac{2}{3} \right) \right)$

$=\tan \left( {{\tan }^{-1}}\left( \frac{3}{4} \right)+{{\tan }^{-1}}\left( \frac{2}{3} \right) \right)$

=$\frac{{}^{3}/{}_{4}+{}^{2}/{}_{3}}{1-\left( \frac{3}{4}\times {}^{2}/{}_{3} \right)}$

$=\frac{17}{6}$ Ans

### Periodicity Of Trigonometric Functions

Periodic function

Functions whose values repeat at regular intervals are called as periodic functions.

Theorem 1

1. sin x is periodic with T = 2π

2. cos x is periodic with T = 2π

3. tan x is periodic with T = π

Theorem 2

f(ax + b) will be of period P/|a| if f(x) is periodic with period T = P

Illustration:

(1) Period of $\sin \left( \frac{3\pi }{2} \right)is\,\frac{2\pi }{(3/2)}=\frac{2\pi }{3}$

(2) 4 + 5 $\cos \left( -3x+\frac{\pi }{4} \right)is\frac{2\pi }{3}$

(3) $\cot \left( \frac{3\pi }{2}+4 \right)is\frac{2\pi }{B}$

Note:

1. If f(x) is periodic with then –f(x) is also periodic with ‘p’.

2. If T is period of f(x) then ‘nT’ is also a period of f(x) but converse need not between

3. If T1 & T2 are [periods pf f1(x) & f2(x) then t(x) = af1(x) + bf2(x) is also periodic with period LCM (T1, T2)

Illustration

1. Find period of tan(x + 2x + 3x + ….nx)

Ans: $f(x)=tan\left( \frac{n(n+1)}{2}x \right)$

∴ $\,T=\frac{\pi }{\left( \frac{n(n+1)}{2} \right)}=\frac{2\pi }{n(n+1)}$

1. Find a cosine function with period 5.

Let cos kx has period 5

∴ $\,\,\,\,\,\frac{2\pi }{|k|}=5\,\,\,\,\,\,\,\,\,\,\,k=\pm \frac{2\pi }{5}$

∴ $\,\,\,\,f(x)=cos\left( \frac{2\pi }{5}x \right)$

1. Find period of function |cos x| f(x) = $\sqrt{{{\cos }^{2}}x}=\sqrt{\frac{1+\cos 2x}{2}}$ as period of cos ax is $\frac{2\pi }{a}$

The period of f(x) is $\frac{2\pi }{2}=\pi$

1. Find period of $\cos x\sin \left( x\frac{-2\pi }{3} \right)$

Ans: $f(x)=\frac{1}{2}\left( 2\cos x\sin \left( x-\frac{2\pi }{3} \right) \right)$

$=\frac{1}{2}\left( \sin \left( 2x-\frac{2\pi }{3} \right)+\sin \left( \frac{-2\pi }{3} \right) \right)$

$=\,\frac{1}{2}\left[ \sin \left( 2x-\frac{2\pi }{3} \right)-\frac{\sqrt{3}}{2} \right].$

(-x) period is $\frac{2\pi }{2}=\pi$ Ans.

### Trigonometric Ratios And Relation Between Sides And Angles Of A Triangle

Let a Δ ABC has side length a, b, & c

S = center of circumcircle

S = Semi perimeter of $\Delta ABC=\frac{a+b+c}{2}$

1. Sine rule in $\Delta ABC=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin c}=2R$
2. if b = 4 cm $\angle A=45{}^\circ \;\;\; \angle B=30{}^\circ$ find value of side BC (a)?

Ans: From sine rule $\frac{a}{\sin 45}=\frac{4}{\sin 30}$ gives $a=4\sqrt{2}$ Ans.

1. If a cosA = b cosB find nature of triangle

Ans: $2{R}$sinA cosA = $2{R}$ sinB cosB

2sin2A = sin 2B

2A = 2B or 2A = 180 – 2B

A = B or A + B = 90°

Either isosceles or right angle triangle.

Theorem 2 cosine Rule

In ΔABC,

b2 = c2 + a2 – 2ac cosB

a2 = b2 + c2 – 2 bc cosA

c2 = O2 + b2 – 2ab cosC

Illustration

1. If a = 2, b = 3, c = 4 find cos A

$\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=\frac{9+16-4}{2\times 3\times 4}=\frac{7}{8}$

1. If b + c: c + a: a + b = 11 : 12 : 13 them prove that

Cos A : cos B : Cos C = 7 : 19 : 25

Ans. b + c : c + a : a + b = 11 : 12 : 13

$\frac{b+c}{11}=\frac{c-a}{12}=\frac{a+b}{13}=k$

b + c = 11k, c + a = 12k, a + b = 13k

a + b + c = 18k

a = 7 k b = 6k c = 5k

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

$\cos \,B=\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ac}=\frac{19}{35}$

$\cos \,C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=\frac{5}{7}$

Hence cosA : cosB : cosC = 7 : 19 : 25

Projection Rule

in Δ ABC,

a = bcos C + c cos B

b = acos C + c cos A

c = a cos B + b cos A

Illustration:

Find, $\frac{b-a\cos c}{c-a\cos B}\,for\,\Delta ABC,$

Ans: $\frac{a\cos C+c\cos A-a\cos C}{a\cos B+b\cos A-a\cos B}=\frac{c}{b}$

HALF ANGLE

1. $\sin {}^{A}/{}_{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}$
2. $\sin {}^{B}/{}_{2}=\sqrt{\frac{(s-c)(s-a)}{ac}}$
3. $\sin {}^{C}/{}_{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}$
4. $\cos {}^{A}/{}_{2}=\sqrt{\frac{s(s-a)}{bc}}$
5. $\cos \left( {}^{B}/{}_{2} \right)=\sqrt{\frac{s(s-b)}{ac}}$
6. $\cos \left( {}^{C}/{}_{2} \right)=\sqrt{\frac{s(s-c)}{ab}}$
7. $\tan \left( {}^{A}/{}_{2} \right)=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
8. $\tan \left( {}^{B}/{}_{2} \right)=\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}$
9. $\tan \left( {}^{C}/{}_{2} \right)=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$
10. $\sin \,A=\frac{2}{bc}\sqrt{s(s-a)(s-b)(s-c)}=\frac{2\Delta }{bc}$
11. $where\,\Delta =\sqrt{s(s-a)(s-b)(s-c)}$
12. $\tan {}^{A}/{}_{2}=\frac{(s-b)(s-c)}{\Delta }=\frac{\Delta }{s(s-a)}\,and\,so\,on$

Area of triangle

1. $\Delta =\frac{1}{2}bc\sin A=\frac{1}{2}ac\sin B=\frac{1}{2}ab\sin C$

$=2{{R}^{2}}\sin A\sin B\sin C$

$=\frac{abc}{4R}$

$=\sqrt{s(s-a)(s-b)(s-c)}$

Illustration:

1. In Δ ABC a = 13, b = 14, c = 15 find tan C/2 = ?

Ans $s=\frac{a+b+c}{2}=21,$

$\tan \left( {}^{c}/{}_{2} \right)=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}={}^{2}/{}_{3}$

2. If a = 5, b = 4 & cos (A – B) = $\frac{31}{32}$ then c =?

Ans c2 = a2 + b2 – 2ab cosc

= 25 + 16 – 2.5.4. cosC

= 41 – 40 cosC

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

$\sqrt{\frac{1-\cos (A-b)}{1+\cos (A+B)}}=\frac{a-b}{a+b}\cot \left( \frac{c}{2} \right)$

3. $\sqrt{\frac{1-\frac{31}{32}}{1-1\frac{31}{32}}}=\frac{5-4}{5+4}\cot \left( {}^{c}/{}_{2} \right)$

⇒ $\,\,\,\tan \left( \frac{C}{2} \right)=\frac{\sqrt{7}}{3}$

⇒ $\,\,\,\cos \,C=\frac{1-{{\tan }^{2}}{}^{C}/{}_{2}}{1-1{{\tan }^{2}}\left( {}^{C}/{}_{2} \right)}=\frac{1-\frac{7}{9}}{1+\frac{7}{9}}=\frac{1}{8}$

⇒ $\,\,\,{{C}^{2}}=41-40\times \frac{1}{8}=36.$

∴ $\,\,\,C=6$

4. show that $\Delta =\frac{{{a}^{2}}}{2(cotB+cotC)}$

Ans: $RHS=\frac{{{a}^{2}}}{2(cotB+cotC)}$

$=\frac{4{{R}^{2}}{{\sin }^{2}}A}{2\left( \frac{\cos B}{\sin B}+\frac{{cosC}}{\sin C} \right)}$

$=\frac{2{{R}^{2}}{{\sin }^{2}}A\sin B\sin C}{\sin (B+C)}$

$=\frac{2{{R}^{2}}{{\sin }^{2}}A\sin B\sin C}{\sin \,A}$

$=2{{R}^{2}}\sin A\sin B\sin C$ = Δ (Ans)