MHB Proving the Relationship between Cosine and Inverse Sine Functions

  • Thread starter Thread starter Wild ownz al
  • Start date Start date
AI Thread Summary
The discussion focuses on proving two relationships involving cosine and inverse sine functions. For part (a), it is established that cos(sin⁻¹x) equals √(1-x²) by applying the identity cos²θ + sin²θ = 1. The proof shows that sin²(sin⁻¹x) equals x², leading to the conclusion that cos(sin⁻¹x) must be the positive square root due to the range of the inverse sine function. In part (b), the equation is clarified to include an equals sign, and the proof involves using the definitions of cos⁻¹a and cos⁻¹b to derive the relationship involving the cosine of the sum of angles. The discussion emphasizes the importance of correctly manipulating the equations and understanding the properties of inverse functions.
Wild ownz al
Messages
30
Reaction score
0
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:
Mathematics news on Phys.org
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.
 
Olinguito said:
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.
 
Wild ownz al said:
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}$

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

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

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

$\cos^{-1}{a} + \cos^{-1}{b} = \cos^{-1}\left[ab - \sqrt{1-a^2} \cdot \sqrt{1-b^2}\right]$
 
Wild ownz al said:
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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top