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Math Help - show identity

  1. #1
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    show identity

    hey im trying to show that sin(X+Y)sin(X-Y) = sin^2X-sin^2Y .

    any idea how i can go with it?

    basically firstly i have:

     <br />
sin(X+Y)sin(X-Y)<br />
= sinXcosY+cosXsinY \cdot sinXcosY-cosXsinY<br />
= <br />

    then where can i go? cheers
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jvignacio View Post
    hey im trying to show that sin(X+Y)sin(X-Y) = sin^2X-sin^2Y .

    any idea how i can go with it?

    basically firstly i have:

     <br />
sin(X+Y)sin(X-Y)<br />
= {\color{red}(}sinXcosY+cosXsinY{\color{red})} \cdot {\color{red}(}sinXcosY-cosXsinY{\color{red})}<br />
= <br />

    then where can i go? cheers
    First, I put needed parentheses into your setup.

    This can be foiled out easily, since it can be recognized as the difference of two squares. You should have

    \begin{aligned}\sin^2X\cos^2Y-\cos^2X\sin^2Y&=\sin^2X\left(1-\sin^2Y\right)-\left(1-\sin^2X\right)\sin^2Y\\&=\sin^2X-\sin^2X\sin^2Y-\sin^2Y+\sin^2X\sin^2Y\\&=\sin^2X-\sin^2Y\end{aligned}

    Does this make sense?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    First, I put needed parentheses into your setup.

    This can be foiled out easily, since it can be recognized as the difference of two squares. You should have

    \begin{aligned}\sin^2X\cos^2Y-\cos^2X\sin^2Y&=\sin^2X\left(1-\sin^2Y\right)-\left(1-\sin^2X\right)\sin^2Y\\&=\sin^2X-\sin^2X\sin^2Y-\sin^2Y+\sin^2X\sin^2Y\\&=\sin^2X-\sin^2Y\end{aligned}

    Does this make sense?
    ahh yes yes that does make sence but quick question. Does cos^2X = (1-sin^2X) ?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jvignacio View Post
    ahh yes yes that does make sence but quick question. Does cos^2X = (1-sin^2X) ?
    Yes, since \sin^2X+\cos^2X=1\implies \cos^2X=1-\sin^2X\text{ or } \sin^2X=1-\cos^2X
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    Yes, since \sin^2X+\cos^2X=1\implies \cos^2X=1-\sin^2X\text{ or } \sin^2X=1-\cos^2X
    of course simple algebra. just gotta know your identities..

    thanks Chris
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    Yes, since \sin^2X+\cos^2X=1\implies \cos^2X=1-\sin^2X\text{ or } \sin^2X=1-\cos^2X
    Hi chris quick question.

    what is (sinXcosY) \cdot (-cosXsinY) ?

    cheers
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jvignacio View Post
    Hi chris quick question.

    what is (sinXcosY) \cdot (-cosXsinY) ?

    cheers
    Its just -\sin X\cos X\sin Y\cos Y.
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