Inverse function of sin , cos and tan

1. Oct 24, 2005

ngkamsengpeter

Solve the following question. Give your answer in cert form .Thank you .
sin-1 x + sin-1 (2x) = pie/3

2. Oct 24, 2005

Tide

Tell us what you have tried so far so we can help you! :)

3. Oct 24, 2005

finchie_88

You could use numerical methods. Example:
f(x) = arcsinx+arcsin2x = pi/3 =>let x(n+1) = sin( pi/3 - arcsinxn), that should work...

4. Oct 24, 2005

Tide

I don't think you have to resort to numerical methods. I'd like to see what the original poster has tried.

5. Oct 26, 2005

ngkamsengpeter

Can you show me the step by step calculations . Thank You .

6. Oct 26, 2005

Tide

If you take the sine of both sides and use the basic properties of inverse trig functions you will find:

$$x \left( \sqrt{1-4x^2} +2 \sqrt {1 - x^2} \right) = \frac {\sqrt 3}{2}$$

I believe you can solve this equation without resorting to numerical approximation.

7. Oct 27, 2005

ngkamsengpeter

I still not understand .Can you explain more detail ? Thank You .

8. Oct 27, 2005

amcavoy

Think about a right triangle. If sin-1(x)=θ, then the hypotenuse must be 1 and the adjacent side $\sqrt{1-x^2}$. If the sine is 2x, then the hypotenuse is still 1, but the cosine (the adjacent side) is $\sqrt{1-4x^2}$. Now try using that to rewrite your equation:

$$\sin{\theta}=x$$

$$\sin{\phi}=2x=2\sin{\theta}$$

9. Oct 27, 2005

Hurkyl

Staff Emeritus
How much do you understand? Have you tried working through the suggested step yourself? Where did you get stuck?

10. Oct 29, 2005

ngkamsengpeter

I do not know what I need to do with the sin-1(x) +sin-1 (2x) .Is it these two have relationship ? If yes please tell me . The suggested step is to short , I cannot understand it . So counld you give me more details on that .

11. Oct 29, 2005

Tide

ng,

When I suggested taking the sine of both sides I assumed you would recognize you need to use a basic trig identity for the sum of angles:

$$\sin a + b = \sin a \cos b + \cos a \sin b$$

Also, you will need basic relations for the inverse trig functions:

$$\sin \sin^{-1} x = x$$

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

and similarly for the cosines. Does that help?

12. Oct 29, 2005

Hurkyl

Staff Emeritus
I was hoping when one poster suggested that you take the sine of both sides, you would have gotten at least as far as

sin( sin-1 x + sin-1 (2x) ) = sin( pie/3 )

13. Oct 30, 2005

ngkamsengpeter

Can you prove for me why $$\sin \cos^{-1} x = \sqrt {1-x^2}$$ . Thank You .

14. Oct 30, 2005

Hurkyl

Staff Emeritus
Have you tried to do it yourself? How far did you get? Have you consulted your book at all?

15. Oct 30, 2005

Tide

I certainly can but I like Hurkyl's questions! Ask again if you get stuck.

16. Oct 30, 2005

amcavoy

You might try starting with the identity:

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

17. Oct 30, 2005

ngkamsengpeter

My book does not teach me that . It only have sin-1 A + cos-1 B = pie/2 . So can you please prove for me because I have no idea about that ?

18. Oct 30, 2005

amcavoy

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

You know that $\cos{\left(\cos^{-1}{x}\right)}=x$, so that above equation becomes:

$$\sqrt{1-x^2}$$

You could also do this geometrically by drawing a right triangle and using the Pythagorean Theorem.

In your last post, did you mean?:

$$\sin^{-1}{x}+\cos^{-1}{x}=\frac{\pi}{2}$$

Last edited: Oct 30, 2005
19. Oct 30, 2005

ngkamsengpeter

Oh yes . I am sorry about the typing error . I have seen your answer but how can we know $$\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}}\right)}$$

20. Oct 30, 2005

Tide

That should be obvious: $\cos \cos^{-1} x = x$ so just square it! :)