# part b help (1 Viewer)

#### ekjchale#1

##### Active Member
part a is ease but wut do i do for part b dont know rlly wut the significance is:

#### Lith_30

##### Active Member
Suppose that $\bg_white f(x)=\cos^{-1}x$ and $\bg_white g(x)=\sin^{-1}\sqrt{1-x^2}$

by part a their derivatives are the same hence $\bg_white f'(x)=g'(x)$

integrating both sides $\bg_white \int{f'(x)}dx=\int{g'(x)}dx\implies{}f(x)=g(x)+c$

now let x = 0
$\bg_white f(0)=\frac{\pi}{2}$

$\bg_white g(0)=\frac{\pi}{2}$

therefore we can say that c=0.

hence $\bg_white \cos^{-1}x=\sin^{-1}\sqrt{1-x^2}\ \text{for}\ 0\leq{x}\leq1$

#### ekjchale#1

##### Active Member
Suppose that $\bg_white f(x)=\cos^{-1}x$ and $\bg_white g(x)=\sin^{-1}\sqrt{1-x^2}$

by part a their derivatives are the same hence $\bg_white f'(x)=g'(x)$

integrating both sides $\bg_white \int{f'(x)}dx=\int{g'(x)}dx\implies{}f(x)=g(x)+c$

now let x = 0
$\bg_white f(0)=\frac{\pi}{2}$

$\bg_white g(0)=\frac{\pi}{2}$

therefore we can say that c=0.

hence $\bg_white \cos^{-1}x=\sin^{-1}\sqrt{1-x^2}\ \text{for}\ 0\leq{x}\leq1$
damn this makes sense now, thanks