MHB Proving the Relationship between Cosine and Inverse Sine Functions

[Wild ownz al]

Prove:

a) cos(sin^-1x) = √(1-x²)

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

(VERY HARD)

 

[Olinguito]

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]

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]

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]

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.