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

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Answer 1 · 2016-08-01T09:47:23

$=(7x)/sqrt(1+49x^2)(sqrt(1-49x^2)-1), x in [-1/7, 1/7]$

Let $a = sin^(-1)(7x) in Q1 or Q4,$ for the principal value..

Then, $sin a = 7x in [-1. 1], and so, x in[-1/7, 1/7].$

and $cos a =sqrt(1-49x^2) in [0, 1]$ .

Let $b = tan^(-1)(7x) in Q1 or Q4$ .

Then, $tan b =7x in[-1, 1] and sin b =(7x)/(1+49x^2)$ .

$cos b = 1/sqrt(49x^2+1) in [0, 1]$ .

Now, the given expressionis

$sin(b-a)$

$=sin b cos a - cos b sin a0$

$=(7x)/sqrt(1+49x^2)sqrt(1-49x^2)-(1/(sqrt(1+49x^2)))(7x)$

$=(7x)/sqrt(1+49x^2)(sqrt(1-49x^2)-1)$

Note that both sines and tangents of a and b are

$>=0$ , when $x>=0$

and are $<=0$ , when $x<=0$ .

Cosines of of both a and b are $>=0$ ..

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