Precalculus

posted by Jess

How would you solve this trig function:

cos2x=sin^2-2


how do you know what trig identities to use?

  1. Damon

    cos 2x = cos^2 x - sin^2 x identity

    so we really have
    cos^2 x - sin^2 x = sin^2 x - 2

    cos^2 x = 2 (sin^2 x-1)
    but sin^2 x = 1 - cos^2 x
    so we have
    cos^2 x = 2 (-cos^2 x)
    or
    3 cos^2 x = 0
    well cos of Pi/2 (90deg) or 3 pi/2 (270 deg) = 0

  2. drwls

    1 - 2sin^2x = sin^2x -2
    3 sin^2x = 3
    sin^2x = 1
    sinx = +1 or -1
    x = pi/2 or 3 pi/2

  3. Jess

    oh i get it.... how do you know what identities to use? did you memorize them?

  4. drwls

    The only ones I have memorized are
    cos 2x = cos^2x - sin^2 x
    sin 2x = 2 sinx cosx,
    and cos^x + sin^2 x = 1.
    I have an old book with many others that I keep handy.

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