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Math Help - Some trig problems =/

  1. #1
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    Some trig problems =/

    I have some work and im confused on these problems... I did all i can (looked in book and such) and need help.

    Simplify the expression cos (X-5pie /2)
    For this one i know that you can go to (cosx)(cos-5pie/2) + (sinx)(sin-5pie/2) But i dont know what to do from there




    second is If cotx= 4/3 and 180degrees< x < 270degrees, find the exact values of sin2x and cos2x
    Here, i know if there was a triangle, there would be the 5,4,and3 to it, but i dont know how you go to the sin2x and cos2x



    third is Find the solutions of sin2x = cosx that are in interval [0, 2pie)

    If anyone could help, youd be my hero. I have 8 tests this week and im stressing out on these problems....
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  2. #2
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    Hello, aargh27!

    1) Simplify: . \cos\left(x - \frac{5\pi}{2}\right)

    What you did was almost correct . . .

    \cos\left(x - \frac{5\pi}{2}\right)\;=\;\cos x\cdot\cos\frac{5\pi}{2} + \sin x\cdot\sin\frac{5\pi}{2}

    Note that \frac{5\pi}{2} is equivalent to \frac{\pi}{2}
    . . so we have: . \cos x\cos\frac{\pi}{2} - \sin x\sin\frac{\pi}{2}

    . . . . . . . . . . = \;\cos x\cdot0 - \sin x\cdot 1 \:=\:\boxed{-\sin x}



    2) If \cot x= \frac{4}{3} and 180^o < x < 270^odegrees,
    find the exact values of \sin2x and \cos2x

    First, note that x is in Quadrant 3.

    Since \cot x = \frac{adj}{opp}, we have: . opp =-3,\;adj = -4\quad\Rightarrow\quad hyp = 5
    . . Then: . \sin x = -\frac{3}{5},\;\cos x = -\frac{4}{5}

    Using double-angle identities, we have:

    . . \sin2x\;=\;2\!\cdot\!\sin x\!\cdot\!\cos x \;=\;2\left(-\frac{3}{5}\right)\left(-\frac{4}{5}\right) \;=\;\boxed{\frac{24}{25}}

    . . \cos2x\;=\;\cos^2x - \sin^2x\;=\;\left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2\;=\;\boxed{\frac{7}{25}}



    3) Find the solutions of: . \sin2x \:= \:\cos x in the interval [0,2\pi)

    Replace \sin2x with 2\sin x\cos x

    . . 2\sin x\cos x \:=\:\cos x\quad\Rightarrow\quad 2\sin x\cos x - \cos x \:=\:0


    Factor: . \cos x(2\sin x - 1)\:=\:0


    And we have two equations to solve:

    . . \cos x\:=\:0\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}
    . . 2\sin x - 1\:=\:0\quad\Rightarrow\quad\sin x = \frac{1}{2}\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{6},\:\frac{5\pi}{6}}

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  3. #3
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    Ty again for your help. However just one thing that I found as these were graded. For the first problem it was +sinx .. not -sinx
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