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Math Help - Solving trig functions...

  1. #1
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    Question Solving trig functions...

    I am having trouble with this problem...It goes as follows:

    Find the exact value of:
    sin(2x), cos(2x), and tan(2x) if cos(x) = -4/5, [pi]/2 <x<[pi].

    It gives me a hint saying( sin2x=2sinxcosx and cos(2x)=2cos^2 (X-1)

    I have no idea how to do this Anyone care to help?
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  2. #2
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    Hello KAY03
    Quote Originally Posted by KAY03 View Post
    I am having trouble with this problem...It goes as follows:

    Find the exact value of:
    sin(2x), cos(2x), and tan(2x) if cos(x) = -4/5, [pi]/2 <x<[pi].

    It gives me a hint saying( sin2x=2sinxcosx and cos(2x)=2cos^2 (X-1)

    I have no idea how to do this Anyone care to help?
    First, note that, if \pi/2 <x<\pi/2, then \sin x is positive.

    Then, using \sin^2x =1-\cos^2x, plug in the value \cos x = -\tfrac45:
    \sin^2x=1 - \tfrac{16}{25}
    =\tfrac{9}{25}
    \Rightarrow\sin x = \tfrac35 (taking the positive square root, since, as we said, \sin x > 0)
    Now you can simply plug in the values \sin x = \tfrac35 and \cos x = -\tfrac45 into the formulae that you've been given. The first one is:
    \sin 2x =2\sin x \cos x
    =2\cdot\tfrac35\cdot(-\tfrac45)

    = ... ?
    Do \cos 2x in the same way, using the formula:
    \cos2x=2\cos^2x -1
    Finally, use \tan2x = \frac{\sin2x}{\cos2x} to find \tan2x.

    Can you complete them now?

    Grandad
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  3. #3
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    reply

    Just to make sure I did it right..my final answers are

    sin2x= -24/25
    cos2x=7/25
    tan2x=-24/7

    Am I correct?
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  4. #4
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    Hello KAY03
    Quote Originally Posted by KAY03 View Post
    Just to make sure I did it right..my final answers are

    sin2x= -24/25
    cos2x=7/25
    tan2x=-24/7

    Am I correct?
    Correct! Good work.

    Grandad
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