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Math Help - Identities

  1. #1
    Newbie
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    Mar 2009
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    Identities

    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 ?

    sin2x
    cos2x
    tan2x

    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
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

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

    Then \sin 2x = 2 \sin x \cos x

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

    \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) \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
    \cos 2x = 1- 2 \sin^2 x

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

    5) \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}
    and A-\sqrt{B})" alt="\cos \frac{\pi}{6} = 1 - 2 \sin^2 \frac{\pi}{12} = 1 - \frac12\A-\sqrt{B})" />
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