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Trigonometry cos(sin^(−1) x) = √(1−x^2)

Wild ownz al

Member
Nov 11, 2018
30
Prove:

a) cos(sin-1x) = √(1-x2)

b) cos-1a+cos-1​b = cos-1(ab-√(1-a2)√(1-b2) (edited)

(VERY HARD)
 
Last edited:

Olinguito

Well-known member
Apr 22, 2018
251
Hi Wild ownz al .

Try using the identity $\cos^2\theta+\sin^2\theta=1$.

(a) We have $\sin^2(\sin^{-1}x)=x^2$ and so
$$\cos^2(\sin^{-1}x)\ =\ 1-\sin^2(\sin^{-1}x)\ =\ 1-x^2$$

$\implies\ \cos(\sin^{-1}x)\ =\ \sqrt{1-x^2}$

taking the positive square root because the range of $\sin^{-1}x$ (for $-1\le x\le1$) is $\displaystyle\left[-\frac{\pi}2,\,\frac{\pi}2\right]$ on which the cos function takes non-negative values.

(b) Check your equation. There should be an equals (“=”) sign, which is missing.
 

Wild ownz al

Member
Nov 11, 2018
30
Hi Wild ownz al .

Try using the identity $\cos^2\theta+\sin^2\theta=1$.

(a) We have $\sin^2(\sin^{-1}x)=x^2$ and so
$$\cos^2(\sin^{-1}x)\ =\ 1-\sin^2(\sin^{-1}x)\ =\ 1-x^2$$

$\implies\ \cos(\sin^{-1}x)\ =\ \sqrt{1-x^2}$

taking the positive square root because the range of $\sin^{-1}x$ (for $-1\le x\le1$) is $\displaystyle\left[-\frac{\pi}2,\,\frac{\pi}2\right]$ on which the cos function takes non-negative values.

(b) Check your equation. There should be an equals (“=”) sign, which is missing.
Hey Olinguito,

I'm a bit confused as to your steps...did you manipulate the left hand side or the right hand side? Also could you start from the given equation? I corrected part b). Thanks.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
644
Prove:

a) cos(sin-1x) = √(1-x2)

b) cos-1a+cos-1​b = cos-1(ab-√(1-a2)√(1-b2) (edited)

(VERY HARD)
(b) let $\theta = \cos^{-1}{a} \implies \cos{\theta} = a \text{ and } \sin{\theta} = \sqrt{1-a^2}$,

also, let $\phi = \cos^{-1}{b} \implies \cos{\phi} = b \text{ and } \sin{\phi} = \sqrt{1-b^2}$


$\cos(\theta + \phi) = \cos{\theta}\cos{\phi} - \sin{\theta}\sin{\phi}$

$\cos(\theta + \phi) = ab - \sqrt{1-a^2} \cdot \sqrt{1-b^2}$

$\cos^{-1}\left[\cos(\theta + \phi)\right] = \cos^{-1}\left[ab - \sqrt{1-a^2} \cdot \sqrt{1-b^2}\right]$

$\theta + \phi = \cos^{-1}\left[ab - \sqrt{1-a^2} \cdot \sqrt{1-b^2}\right]$

$\cos^{-1}{a} + \cos^{-1}{b} = \cos^{-1}\left[ab - \sqrt{1-a^2} \cdot \sqrt{1-b^2}\right]$
 

Olinguito

Well-known member
Apr 22, 2018
251
I'm a bit confused as to your steps...did you manipulate the left hand side or the right hand side? Also could you start from the given equation? I corrected part b). Thanks.
Note that the $\sin$ and $\sin^{-1}$ are inverse functions.

For example, $\sin^{-1}\dfrac12=\dfrac{\pi}6$ and $\sin\dfrac{\pi}6=\dfrac12$; that is to say, $\sin\left(\sin^{-1}\dfrac12\right)=\dfrac12$.

Thus we have $\sin(\sin^{-1}x)=x$ for $-1\le x\le1$.

The rest of my post should be straightforward.