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Thread: Identities

  1. #1
    Mar 2009


    It's been a while since my last math class, and my current teacher isnt the greatest. We're assigned a fair amount of homework everynight, and it just takes me forever to do it, and its taking all of my time, leaving no time left for other classes. I have some problems and if anyone can help me with them, I'd greatly appreciate it. I learn best by seeing an example, but our teacher tends not to give many examples, if any.

    2) If sin x = −3/5,x in quadrant III, then what are the following ?


    4)If sin4 x = A+Bcos2x+Ccos4x, then what are A, B, C ?

    5If sin(15Degrees) = 1/2 sqrt(A - sqrt(B))
    B, then, by using a half-angle formula, find A , B

    I have some more, but help with these would be amazing.
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  2. #2
    MHF Contributor
    Nov 2008

    2) Knowing $\displaystyle \sin x$ you can compute $\displaystyle \cos^2x = 1 - \sin^2 x$
    Then x being in quadrant III, you can compute $\displaystyle \cos x$

    Then $\displaystyle \sin 2x = 2 \sin x \cos x$

    $\displaystyle \cos 2x = 1- 2 \sin^2 x$

    $\displaystyle \tan 2x = \frac{\sin 2x}{\cos 2x}$

    4) 2 possibilities (at least)

    a) Chose some values (x=0, x=pi/2, ...) to get 3 linear equations with unknowns A, B and C

    b) $\displaystyle \cos 4x = \cos 2(2x) = 2 \cos^2 2x -1 = 2(1-2\sin^2 x)^2-1 = 8 \sin^4 x - 8 \sin^2 x + 1$
    $\displaystyle \cos 2x = 1- 2 \sin^2 x$

    Therefore you can find the expression of $\displaystyle A + B \cos 2x + C \cos 4x$ and find A, B and C such that $\displaystyle A + B \cos 2x + C \cos 4x = \sin^4 x$

    5) $\displaystyle \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$
    and $\displaystyle \cos \frac{\pi}{6} = 1 - 2 \sin^2 \frac{\pi}{12} = 1 - \frac12\A-\sqrt{B})$
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