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Math Help - Losing my mind

  1. #1
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    Losing my mind

    I have searched my book front to back to find just one example of this type of problem and fond nothing. I
    have also looked through both this forum and yahoo answers and saw nothing this ridiculous.

    PLZ help me and save me from an anurism

    Sin x = -2/5 x in quad 3

    exact value of cos (x+Pi/4)




    and the second seems easier but im afraid to try it

    Exact value of sin 2x

    I already did something i though helped, but got me no closer to the answer

    sin x^2 + cos x^2 = 1
    cos x^2 = 1+ 2/5^2=1 + 4/25=29/25=sqrt-29/25=sqrt-29/5


    and this is where i contemplate my lifes worth..... This problem was on my final at OSU and i need a good grade! Help!
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    North Texas
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    x in quad III

    \sin{x} = -\frac{2}{5}

    sketch a reference triangle in quad III ... opposite side = -2, hypotenuse = 5

    adjacent side = -\sqrt{5^2 - 2^2} = -\sqrt{21}

    \cos{x} = -\frac{\sqrt{21}}{5}

    now that you have the necessary values ...

    \cos\left(x + \frac{\pi}{4}\right) = \cos{x}\cos\left(\frac{\pi}{4}\right) - \sin{x}\sin\left(\frac{\pi}{4}\right)

    \sin(2x) = 2\sin{x}\cos{x}
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