# 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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