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Math Help - Another Trig Identity problem.

  1. #1
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    Another Trig Identity problem.

    I need to know in steps how to prove this identity, if anyone minds helping. Thanks!

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  2. #2
    MHF Contributor harish21's Avatar
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    \tan(x)=\dfrac{\sin(x)}{\cos(x)}

    \therefore \cos(x)\sin(x)\tan(x)=\cos(x)\sin(x)\dfrac{\sin(x)  }{\cos(x)}=....
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  3. #3
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    Quote Originally Posted by harish21 View Post
    \tan(x)=\dfrac{\sin(x)}{\cos(x)}

    \therefore \cos(x)\sin(x)\tan(x)=\cos(x)\sin(x)\dfrac{\sin(x)  }{\cos(x)}=....
    Can't solve?
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  4. #4
    MHF Contributor harish21's Avatar
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    Are you even attempting?

    You should be able to recognize that the cos(x) in the numerator and the denominator cancel each other out. So you are left with \cos(x)\sin(x)\dfrac{\sin(x)}{\cos(x)} = \sin(x)\sin(x)=\sin^2(x)

    You have been told in your previous post that sin^2(x)+cos^2(x)=1. based on that, \sin^2(x)= ??
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  5. #5
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    I didn't realize they canceled each other out, but when I did the equation myself I canceled them out anyway. When I was left with I just added the negative to the other side to make a positive, thus making it all equal one. I thought I was doing it wrong though, since I didn't know they canceled. Thanks for clarification.
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