Simplify Inverse Trig Function

**krizzle13** · December 2nd 2008, 01:28 AM

Again, i had mono and got behind...thank you

Smplify the expression for f(x):

1. f(x) = sin(cos^-1(x) + sin^-1(x))

2. f(x) = tan(pi + sin^-1(x))

sorry, i am not good at typing math...thanks for all your help

**Chop Suey** · December 2nd 2008, 02:43 AM

1. Easily done using the identities:
$\sin{(A \pm B)} = \sin{A}\cos{B} \pm \sin{B}\cos{A}$

$\sin^2{\theta}+\cos^2{\theta} = 1$

============

Let $\alpha = \arccos{x} \implies \cos{\alpha}=x$ and let $\beta = \arcsin{x} \implies \sin{\beta}=x$

$\sin{(\alpha + \beta)} = \sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha}$

$= \sqrt{1-\cos^2{\alpha}}\sqrt{1-\sin^2{\beta}}+\sin{\beta}\cos{\alpha}$

$= \sqrt{1-x^2}\sqrt{1-x^2}+x^2$

$= 1$

2. Recall that $-\tan{x} = \tan{(\pi-x)}$ and $\tan^2{x}+1=\sec^2{x}$

==============

$\tan{(\pi-(-\arcsin{x}))} = -\tan{(-\arcsin{x})}=\tan{(\arcsin{x})}$

Let $\alpha = \arcsin{x} \implies \sin{\alpha} = x$

$\tan{\alpha} = \sqrt{\frac{1}{\cos^2{x}}-1} = \sqrt{\frac{1}{1-\sin^2{\alpha}}-1}$

Simply replace and you're done.

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