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Math Help - Verifying Trigonometric Identities

  1. #1
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    Verifying Trigonometric Identities

    I'm a bit confused on verifying trigonometric identities. An example problem would be:

    \cos(\frac{\pi}{2})-\Theta=\sin\Theta

    Or:

    \cot(\pi+\Theta)=\cot\Theta

    Can someone please explain what exactly the question is asking, and how I find the solution? Thanks.

    Also, is there any way to visually differ between the graph of a sine function and the graph of a cosine function, if you don't know if one may be shifted? Sorry if that is hard to understand.
    Last edited by AlderDragon; May 4th 2009 at 10:32 PM.
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  2. #2
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    The question is asking to show that the expressions of LHS and RHS are equal, i.e. that you can use some identities to show equality.
    Take the first one:

    \cos(\frac{\pi}{2}-\theta)=<br />
\cos(\frac{\pi}{2})\cos\theta+\sin(\frac{\pi}{2})\  sin\theta<br />
=\sin\theta.

    For the graph question. I distinct between the sine and cosine graphs by remembering that sine passes through the origin, while cosine does not pass through the origin.
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  3. #3
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    Quote Originally Posted by vemrygh View Post
    The question is asking to show that the expressions of LHS and RHS are equal, i.e. that you can use some identities to show equality.
    Take the first one:

    \cos(\frac{\pi}{2}-\theta)=<br />
\cos(\frac{\pi}{2})\cos\theta+\sin(\frac{\pi}{2})\  sin\theta<br />
=\sin\theta.

    For the graph question. I distinct between the sine and cosine graphs by remembering that sine passes through the origin, while cosine does not pass through the origin.





    I still don't understand how to solve those problems. For:
    <br /> <br />
\cos(\frac{\pi}{2}-\theta)=<br />
\cos(\frac{\pi}{2})\cos\theta+\sin(\frac{\pi}{2})\  sin\theta<br />
=\sin\theta<br />

    How do you get from the original problem to:

    <br />
\cos(\frac{\pi}{2}-\theta)\cos\Theta<br />

    And from that to:

    <br />
\cos(\frac{\pi}{2})\cos\theta+\sin(\frac{\pi}{2})\  sin\theta<br />
    Last edited by AlderDragon; May 5th 2009 at 06:25 AM.
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  4. #4
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    Quote Originally Posted by AlderDragon View Post





    I still don't understand how to solve those problems. For:
    <br /> <br />
\cos(\frac{\pi}{2}-\theta)=<br />
\cos(\frac{\pi}{2})\cos\theta+\sin(\frac{\pi}{2})\  sin\theta<br />
=\sin\theta<br />

    How do you get from the original problem to:

    <br />
\cos(\frac{\pi}{2}-\theta)\cos\Theta<br />

    And from that to:

    <br />
\cos(\frac{\pi}{2})\cos\theta+\sin(\frac{\pi}{2})\  sin\theta<br />
    Hi

    vemrygh used the trig identity cos(a-b)=cos a cos b + sin a sin b
    with a = pi/2 and b = theta
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