Inverse function of sin , cos and tan

[ngkamsengpeter]

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

 

[Tide]

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

 

[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...

 

[Tide]

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

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

 

[ngkamsengpeter]

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

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

 

[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.

 

[ngkamsengpeter]

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.

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

 

[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}

 

[Hurkyl]

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

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

 

[ngkamsengpeter]

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

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 .

 

[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?

 

[Hurkyl]

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 )

 

[ngkamsengpeter]

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?

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

 

[Hurkyl]

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

 

[Tide]

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

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

 

[amcavoy]

You might try starting with the identity:

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

 

[ngkamsengpeter]

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

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 ?

 

[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}

 

[ngkamsengpeter]

\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}

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)}

 

[Tide]

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

 

[amcavoy]

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)}

Look at post #16.

 

[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)}

Based on my knowledge, \cos^2{\left(\cos^{-1}{x}\right)} should become cos x right ? How can cos^2{\left(\cos^{-1}{x}}\right)} become {x}^2 ?

 

[ngkamsengpeter]

Based on my knowledge, \cos^2{\left(\cos^{-1}{x}\right)} should become cos x right ? How can cos^2{\left(\cos^{-1}{x}}\right)} become {x}^2 ?

Someone please answer my question . Thank you .

 

[Tide]

Break it down ...
\cos ^2 (\cos^{-1} x) = \left( \cos \cos^{-1} x\right) \times \left( \cos \cos^{-1} x\right)