part b help (1 Viewer)

ekjchale#1 · Jun 7, 2022

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

Lith_30 · Jun 7, 2022

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

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

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

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

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

therefore we can say that c=0.

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

ekjchale#1 · Jun 7, 2022

Lith_30 said:
Suppose that $f(x)=\cos^{-1}x$ and $g(x)=\sin^{-1}\sqrt{1-x^2}$

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

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

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

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

therefore we can say that c=0.

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

damn this makes sense now, thanks

part b help (1 Viewer)

ekjchale#1

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Lith_30

o_o

ekjchale#1

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