Posted by **Jess** on Saturday, August 6, 2011 at 5:39pm.

How would you solve this trig function:

cos2x=sin^2-2

how do you know what trig identities to use?

- Precalculus -
**Damon**, Saturday, August 6, 2011 at 5:47pm
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

- Precalculus -
**drwls**, Saturday, August 6, 2011 at 5:51pm
1 - 2sin^2x = sin^2x -2

3 sin^2x = 3

sin^2x = 1

sinx = +1 or -1

x = pi/2 or 3 pi/2

- Precalculus -
**Jess**, Saturday, August 6, 2011 at 6:04pm
oh i get it.... how do you know what identities to use? did you memorize them?

- Precalculus -
**drwls**, Saturday, August 6, 2011 at 11:13pm
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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