Inverse function of sin , cos and tan

In summary: Oh yes . I am sorry about the typing error . I have seen your answer but how can we know...\sin^{-1}{x}+\cos^{-1}{x}=\frac{\pi}{2}\)
  • #1
ngkamsengpeter
195
0
Solve the following question. Give your answer in cert form .Thank you .
sin-1 x + sin-1 (2x) = pie/3
 
Physics news on Phys.org
  • #2
Tell us what you have tried so far so we can help you! :)
 
  • #3
You could use numerical methods. Example:
f(x) = arcsinx+arcsin2x = pi/3 =>let x(n+1) = sin( pi/3 - arcsinxn), that should work...
 
  • #4
finchie_88 said:
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.
 
  • #5
finchie_88 said:
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 .
 
  • #6
If you take the sine of both sides and use the basic properties of inverse trig functions you will find:

[tex]x \left( \sqrt{1-4x^2} +2 \sqrt {1 - x^2} \right) = \frac {\sqrt 3}{2}[/tex]

I believe you can solve this equation without resorting to numerical approximation.
 
  • #7
Tide said:
If you take the sine of both sides and use the basic properties of inverse trig functions you will find:
[tex]x \left( \sqrt{1-4x^2} +2 \sqrt {1 - x^2} \right) = \frac {\sqrt 3}{2}[/tex]
I believe you can solve this equation without resorting to numerical approximation.

I still not understand .Can you explain more detail ? Thank You .
 
  • #8
Think about a right triangle. If sin-1(x)=θ, then the hypotenuse must be 1 and the adjacent side [itex]\sqrt{1-x^2}[/itex]. If the sine is 2x, then the hypotenuse is still 1, but the cosine (the adjacent side) is [itex]\sqrt{1-4x^2}[/itex]. Now try using that to rewrite your equation:

[tex]\sin{\theta}=x[/tex]

[tex]\sin{\phi}=2x=2\sin{\theta}[/tex]
 
  • #9
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?
 
  • #10
Hurkyl said:
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 .
 
  • #11
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:

[tex]\sin a + b = \sin a \cos b + \cos a \sin b[/tex]

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

[tex]\sin \sin^{-1} x = x[/tex]

[tex]\sin \cos^{-1} x = \sqrt {1-x^2}[/tex]

and similarly for the cosines. Does that help?
 
  • #12
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
Tide said:
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:
[tex]\sin a + b = \sin a \cos b + \cos a \sin b[/tex]
Also, you will need basic relations for the inverse trig functions:
[tex]\sin \sin^{-1} x = x[/tex]
[tex]\sin \cos^{-1} x = \sqrt {1-x^2}[/tex]
and similarly for the cosines. Does that help?
Can you prove for me why [tex]\sin \cos^{-1} x = \sqrt {1-x^2}[/tex] . Thank You .
 
  • #14
Have you tried to do it yourself? How far did you get? Have you consulted your book at all?
 
  • #15
ngkamsengpeter said:
Can you prove for me why [tex]\sin \cos^{-1} x = \sqrt {1-x^2}[/tex] . Thank You .

I certainly can but I like Hurkyl's questions! Ask again if you get stuck. :smile:
 
  • #16
You might try starting with the identity:

[tex]\sin{x}=\sqrt{1-\cos^2{x}}[/tex]
 
  • #17
Hurkyl said:
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 ?
 
  • #18
[tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}}\right)}[/tex]

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

[tex]\sqrt{1-x^2}[/tex]

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

In your last post, did you mean?:

[tex]\sin^{-1}{x}+\cos^{-1}{x}=\frac{\pi}{2}[/tex]
 
Last edited:
  • #19
apmcavoy said:
[tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}}\right)}[/tex]
You know that [itex]\cos{\left(\cos^{-1}{x}\right)}=x[/itex], so that above equation becomes:
[tex]\sqrt{1-x^2}[/tex]
You could also do this geometrically by drawing a right triangle and using the Pythagorean Theorem.
In your last post, did you mean?:
[tex]\sin^{-1}{x}+\cos^{-1}{x}=\frac{\pi}{2}[/tex]
Oh yes . I am sorry about the typing error . I have seen your answer but how can we know [tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}}\right)}[/tex]
 
  • #20
That should be obvious: [itex]\cos \cos^{-1} x = x[/itex] so just square it! :)
 
  • #21
ngkamsengpeter said:
Oh yes . I am sorry about the typing error . I have seen your answer but how can we know [tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}}\right)}[/tex]
Look at post #16.
 
  • #22
ngkamsengpeter said:
Oh yes . I am sorry about the typing error . I have seen your answer but how can we know [tex]\sin{\left(\cos^{-1}{x}\right)}=\sqrt{1-\cos^2{\left(\cos^{-1}{x}\right)}[/tex]
Based on my knowledge, [tex]\cos^2{\left(\cos^{-1}{x}\right)} [/tex] should become [tex]cos x[/tex] right ? How can [tex]cos^2{\left(\cos^{-1}{x}}\right)} [/tex] become [tex]{x}^2[/tex] ?:confused:
 
Last edited:
  • #23
ngkamsengpeter said:
Based on my knowledge, [tex]\cos^2{\left(\cos^{-1}{x}\right)} [/tex] should become [tex]cos x[/tex] right ? How can [tex]cos^2{\left(\cos^{-1}{x}}\right)} [/tex] become [tex]{x}^2[/tex] ?:confused:
Someone please answer my question . Thank you .:frown:
 
  • #24
Break it down ...
[tex]\cos ^2 (\cos^{-1} x) = \left( \cos \cos^{-1} x\right) \times \left( \cos \cos^{-1} x\right)[/tex]
 

What is the inverse function of sin?

The inverse function of sin is called arcsin or sin-1. It is the function that undoes the output of the sin function, and is defined as the angle whose sine is equal to a given value.

What is the inverse function of cos?

The inverse function of cos is called arccos or cos-1. It is the function that undoes the output of the cos function, and is defined as the angle whose cosine is equal to a given value.

What is the inverse function of tan?

The inverse function of tan is called arctan or tan-1. It is the function that undoes the output of the tan function, and is defined as the angle whose tangent is equal to a given value.

What is the domain and range of inverse trigonometric functions?

The domain of inverse trigonometric functions is the set of values that can be plugged into the function to produce a valid output. The range is the set of all possible outputs of the function. For inverse trigonometric functions, the domain is usually restricted to a specific interval in order to ensure a unique output.

How are inverse trigonometric functions related to their corresponding trigonometric functions?

The inverse trigonometric functions are essentially the inverse operations of their corresponding trigonometric functions. For example, arcsin undoes the output of sin, arccos undoes the output of cos, and arctan undoes the output of tan. They can also be thought of as the "undo" buttons for trigonometric functions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
236
  • Calculus and Beyond Homework Help
Replies
2
Views
836
  • Calculus and Beyond Homework Help
Replies
15
Views
735
  • Calculus and Beyond Homework Help
Replies
10
Views
897
  • Calculus and Beyond Homework Help
Replies
11
Views
196
  • Calculus and Beyond Homework Help
Replies
1
Views
55
  • Calculus and Beyond Homework Help
Replies
2
Views
782
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
28
Views
1K
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
Back
Top