Results 1 to 3 of 3

Thread: inverse trig squaring

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    40

    inverse trig squaring

    By squaring both sides of y=sincos^-1 (x) and using the identity sin^2 x + cos^2 x = 1. Show that y=Sqrt(1-x^2)

    I'm not sure how to square an inverse function.

    Thannk you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by deltaxray View Post
    By squaring both sides of y=sincos^-1 (x) and using the identity sin^2 x + cos^2 x = 1. Show that y=Sqrt(1-x^2)

    I'm not sure how to square an inverse function.

    Thannk you
    Whenever you have an inverse function, you should give it a name. That helps. Let us call $\displaystyle \cos^{-1} (x)$ as $\displaystyle \theta$. This means $\displaystyle \cos \theta = x$.
    Now let us read your question. You know that $\displaystyle y = \sin\cos^{-1} x = \sin \theta$

    Now can you write $\displaystyle \sin \theta$ in terms of $\displaystyle \cos \theta$ and finish the problem?

    Do not forget to substitute the variable $\displaystyle \theta$ back.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    40
    thank you

    I got it
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Feb 24th 2011, 09:48 AM
  2. Find the value of 'trig of inverse trig' questions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Apr 9th 2010, 05:37 PM
  3. inverse trig.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Nov 4th 2009, 02:46 AM
  4. inverse trig values and finding inverse
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Apr 6th 2009, 12:04 AM
  5. Inverse Trig Functions and Advance Trig
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Sep 24th 2008, 03:13 PM

Search Tags


/mathhelpforum @mathhelpforum