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Thread: Losing my mind

  1. #1
    Dec 2008

    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
    Jun 2008
    North Texas
    x in quad III

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

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

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

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

    now that you have the necessary values ...

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

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