# How do you simplify sin(tan^-1(7x)-sin^-1(7x))?

Aug 1, 2016

$= \frac{7 x}{\sqrt{1 + 49 {x}^{2}}} \left(\sqrt{1 - 49 {x}^{2}} - 1\right) , x \in \left[- \frac{1}{7} , \frac{1}{7}\right]$

#### Explanation:

Let $a = {\sin}^{- 1} \left(7 x\right) \in Q 1 \mathmr{and} Q 4 ,$ for the principal value..

Then, $\sin a = 7 x \in \left[- 1. 1\right] , \mathmr{and} s o , x \in \left[- \frac{1}{7} , \frac{1}{7}\right] .$

and $\cos a = \sqrt{1 - 49 {x}^{2}} \in \left[0 , 1\right]$.

Let $b = {\tan}^{- 1} \left(7 x\right) \in Q 1 \mathmr{and} Q 4$.

Then, $\tan b = 7 x \in \left[- 1 , 1\right] \mathmr{and} \sin b = \frac{7 x}{1 + 49 {x}^{2}}$.

$\cos b = \frac{1}{\sqrt{49 {x}^{2} + 1}} \in \left[0 , 1\right]$.

Now, the given expressionis

$\sin \left(b - a\right)$

$= \sin b \cos a - \cos b \sin a 0$

$= \frac{7 x}{\sqrt{1 + 49 {x}^{2}}} \sqrt{1 - 49 {x}^{2}} - \left(\frac{1}{\sqrt{1 + 49 {x}^{2}}}\right) \left(7 x\right)$

$= \frac{7 x}{\sqrt{1 + 49 {x}^{2}}} \left(\sqrt{1 - 49 {x}^{2}} - 1\right)$

Note that both sines and tangents of a and b are

$\ge 0$, when $x \ge 0$

and are $\le 0$, when $x \le 0$.

Cosines of of both a and b are $\ge 0$..