How do I prove that #(3-6 cos^2 x)/(sin x- cos x)# = 3(sin x + cos x)?

1 Answer
Sep 20, 2015

Answer:

Yes, the equality is correct. See explanation.

Explanation:

You have
#frac(3-6cos^2 x)(sin x - cos x)#

On the numerator, factor out 3:
#= 3*frac(1-2cos^2 x)(sin x - cos x)#

Remember the identity #sin^2 x + cos^2 x = 1#.
Substitute the 1 in the numerator:
#= 3*frac(sin^2 x + cos^2 x -2cos^2 x)(sin x - cos x)#

Simplify by adding the cos^2 x together:
#= 3*frac(sin^2 x -cos^2 x)(sin x - cos x)#

Remember that when you have something of the form #a^2 - b^2 = (a-b)(a+b)#

So rewrite the numerator
#= 3*frac((sin x -cos x)(sin x + cos x))(sin x - cos x)#

And remove #sin x - cos x# from the numerator and denominator. You are left with:

#=3(sin x + cos x)#

Q.E.D.