How would you solve this trig function:
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)
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.