# Inverse Trigonometric Functions

Let’s keep it simple and straight. The quantities such as $$\sin{}^{-1}\alpha$$, $$\cos{}^{-1}\alpha$$, $$\tan{}^{-1}\alpha$$, … are Inverse Trigonometric Functions. We also call them Inverse Circular or Cyclometric Functions. These are widely used inter alia in engineering and navigation.

So, $$\sin{}^{-1}\alpha$$ is an angle. Precisely, it denotes the smallest angle whose sine is $$\alpha$$. That is, if $$\sin\theta = \alpha$$, then $$\sin{}^{-1}\alpha$$ is equal to the smallest angle $$(\theta)$$ that satisfies the relation. The angle may be positive or negative.

We know that for $$\sin\theta = \alpha$$ where $$\alpha$$ is a known definite quantity, $$(\theta)$$ is not a known definite angle but one of a series of angles.

To note,

• $$\sin{}^{-1}\alpha$$, $$\tan{}^{-1}\alpha$$, $$\csc{}^{-1}\alpha$$ and $$\cot{}^{-1}\alpha$$ always lie between $$-90{°}$$ and $$+90{°}$$;
• when $$\alpha$$ is $$+$$ve, $$\sin{}^{-1}\alpha$$ lies between $$0{°}$$ and $$+90{°}$$;
• when $$\alpha$$ is $$-$$ve, $$\sin{}^{-1}\alpha$$ lies between $$-90{°}$$ and $$0{°}$$; and
• $$\cos{}^{-1}\alpha$$ and $$\sec{}^{-1}\alpha$$ invariably lie between $$0{°}$$ and $$180{°}$$.

It should be clear by now that $$\sin{}^{-1}\alpha$$ is not equivalent to $$\frac{1}{\sin(\alpha)}$$. In fact, if we wish, we may write the later as $$(\sin\alpha)^{-1}$$. Moreover, we also write $$\sin{}^{-1}\alpha$$, $$\cos{}^{-1}\alpha$$, $$\tan{}^{-1}\alpha$$, … as $$\arcsin(x)$$, $$\arccos(x)$$, $$\arctan(x)$$, …

## SL Loney – Examples 43

Prove these:

1. $$\sin{}^{-1}\frac{3}{5} + \sin{}^{-1}\frac{8}{17} = \sin{}^{-1}\frac{77}{85}$$

3. $$\cos{}^{-1}\frac{4}{5} + \tan{}^{-1}\frac{3}{5} = \tan {}^{-1}\frac{27}{11}$$

7. $$\tan {}^{-1}\frac{1}{2} + \tan {}^{-1}\frac{1}{3} = 45{°}$$

8. $$\tan {}^{-1}\frac{1}{7} + \tan {}^{-1}\frac{1}{13} = \tan {}^{-1}\frac{2}{9}$$

9. $$2\tan{}^{-1}\frac{2}{3} = \tan{}^{-1}\frac{12}{5}$$

13. $$\tan{}^{-1}\frac{1}{3} + \tan{}^{-1}\frac{1}{5} + \tan{}^{-1}\frac{1}{7} + \tan{}^{-1}\frac{1}{8} = \frac{\pi}{4}$$

17. $$\tan{}^{-1}\frac{m}{n} – \tan{}^{-1}\frac{m – n}{m + n} = \frac{\pi}{4}$$

39. Solve: $$\sin {}^{-1}x + \sin {}^{-1}2x = \frac{\pi}{3}$$

## Solutions

#1.
Let $$\sin{}^{-1}\frac{3}{5} = A$$
$$\Rightarrow \sin A = \frac{3}{5}$$ and $$\cos A = \frac{4}{5}$$

Similarly, for 2nd term:
$$\sin B = \frac{8}{17}$$ and $$\cos B = \frac{15}{17}$$

$$\therefore \sin(A + B) = \sin A \cos B + \cos A \sin B = \frac{77}{85}$$

$$\Rightarrow A + B = \sin{}^{-1}\frac{77}{85}$$

Hence, $$\sin{}^{-1}\frac{3}{5} + \sin{}^{-1}\frac{8}{17} = \sin{}^{-1}\frac{77}{85}$$

#3.
Consider LHS:
Let 1st term $$= \cos \alpha$$ and 2nd term $$= \tan \beta$$
then, $$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – tan\alpha \cdot tan\beta} = \frac{27}{11}$$

or, $$\alpha + \beta = \tan {}^{-1}\frac{27}{11}$$
Hence, $$\cos{}^{-1}\frac{4}{5} + \tan{}^{-1}\frac{3}{5} = \tan {}^{-1}\frac{27}{11}$$

#7.
Let $$\tan {}^{-1}\frac{1}{2} = \alpha$$ and $$\tan {}^{-1}\frac{1}{3} = \beta$$
then, $$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – tan\alpha \cdot tan\beta} = 1 = \tan 45{°}$$

Hence, $$\tan {}^{-1}\frac{1}{2} + \tan {}^{-1}\frac{1}{3} = 45{°}$$

#8.
Let $\tan {}^{-1}\frac{1}{7} = \alpha \Rightarrow \tan \alpha = \frac{1}{7}$ $\tan {}^{-1}\frac{1}{13} = \beta \Rightarrow \tan \beta = \frac{1}{13}$ $\tan {}^{-1}\frac{2}{9} = \gamma \Rightarrow \tan \gamma = \frac{2}{9}$ Now, $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – \tan(\alpha)\tan(\beta)} = \frac{\frac{1}{7} + \frac{1}{13}}{1 – (\frac{1}{7})(\frac{1}{13})} = \frac{2}{9}$ $\Rightarrow (\alpha + \beta) = \tan {}^{-1}\frac{2}{9}$ $\Rightarrow \tan {}^{-1}\frac{1}{7} + \tan {}^{-1}\frac{1}{13} = \tan {}^{-1}\frac{2}{9} = RHS$

#9.
Let $$\tan{}^{-1}\frac{2}{3} = \alpha \Rightarrow \tan(\alpha) = \frac{2}{3}$$

Now, $$\tan(2\alpha) = \frac{2\tan(\alpha)}{1 – \tan(\alpha)^2} = \frac{12}{9}$$
$$\therefore 2\alpha = \tan{}^{-1}\frac{12}{9}$$

Or, $$2\tan{}^{-1}\frac{2}{3} = \tan{}^{-1}\frac{12}{5}$$

#13.
Here, LHS $$= \tan{}^{-1}\frac{\frac{1}{3} + \frac{1}{5}}{1 – \frac{1}{3} \cdot \frac{1}{5}} + \tan{}^{-1}\frac{\frac{1}{7} + \frac{1}{8}}{1 – \frac{1}{7} \cdot \frac{1}{8}}$$

$$= \tan{}^{-1}\frac{4}{7} + \tan{}^{-1}\frac{3}{11} = \tan{}^{-1}\frac{\frac{4}{7} + \frac{3}{11}}{1 – \frac{4}{7} \cdot \frac{3}{11}}$$

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

#17.
LHS$$= \tan{}^{-1}\frac{\frac{m}{n} – \frac{m – n}{m + n}}{1 + \frac{m}{n} \cdot \frac{m – n}{m + n}} = \tan{}^{-1}1 = \frac{\pi}{4} =$$ RHS

#39.
Let $\sin{}^{-1}x = \alpha$ then, $\sin\alpha = x$ and $\cos\alpha = \sqrt{1 – x^2}$Similarly, from 2nd term $\sin\beta = 2x$ and $\cos\beta = \sqrt{1 – 4x^2}$so, the eqn. becomes$\alpha + \beta = \frac{\pi}{3}$or, $\cos(\alpha + \beta) = \cos\frac{\pi}{3}$$\Rightarrow \cos\alpha\cos\beta – \sin\alpha\sin\beta = \frac{1}{2}$Putting values,$x = \pm \frac{1}{2}\sqrt{\frac{3}{7}}$