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Thread: Trig. Identities

  1. #1
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    Trig. Identities

    Hi all,

    I am revising maths trig I am trying to work out this.

    Prove; $\displaystyle cos[theta] = 1-sin[alpha]tan([theta]/2)$

    [theta] is alpha symbol.

    I have tried but keep ending up with $\displaystyle sin^2([theta]/2) + sin^2([theta]/2)$

    Thanks.
    Last edited by Googl; Feb 9th 2011 at 07:11 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Googl View Post
    Hi all,

    I am revising maths trig I am trying to work out this.

    Prove; $\displaystyle cos[alpha] = 1-sinθtan([alpha]/2)$

    [alpha] is alpha symbol.

    I have tried but keep ending up with $\displaystyle sin^2([alpha]/2) + sin^2([alpha]/2)$

    Thanks.
    What is "sin(952)"?

    and what does the tan function after it indicate?

    -Dan
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  3. #3
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    Hello, Googl!

    You should be familiar with these identities:

    . . $\displaystyle \cos\theta \:=\:1-2\sin^2\frac{\theta}{2}$

    . . $\displaystyle \sin\theta \:=\:2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$



    $\displaystyle \text{Prove: }\:\cos\theta \:=\: 1- \sin\theta\tan\frac{\theta}{2}$

    $\displaystyle \text{The right side is: }\:1 - \left(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\ri ght)\left(\dfrac{\sin\frac{\theta}{2}}{\cos\frac{\ theta}{2}}\right) $

    . . . . . . . . . . . $\displaystyle =\;1 - 2\sin^2\frac{\theta}{2}$

    . . . . . . . . . . . $\displaystyle =\;\cos\theta $

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  4. #4
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    Quote Originally Posted by topsquark View Post
    What is "sin(952)"?

    and what does the tan function after it indicate?

    -Dan
    Sorry that is supposed to be theta, I type it but does not come out right.
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Googl!

    You should be familiar with these identities:

    . . $\displaystyle \cos\theta \:=\:1-2\sin^2\frac{\theta}{2}$

    . . $\displaystyle \sin\theta \:=\:2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$




    $\displaystyle \text{The right side is: }\:1 - \left(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\ri ght)\left(\dfrac{\sin\frac{\theta}{2}}{\cos\frac{\ theta}{2}}\right) $

    . . . . . . . . . . . $\displaystyle =\;1 - 2\sin^2\frac{\theta}{2}$

    . . . . . . . . . . . $\displaystyle =\;\cos\theta $

    Thanks.

    I think I got to the identity 2sin(theta/2)cos(theta/2) but did not realise it was equal to $\displaystyle sin\theta$
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