How do you use the double angle formula to rewrite #5-10sin^2x#?

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Answer 1 · 2015-04-26T17:32:59

Think about (list) double angle formulas until you come to one that looks useful.

#sin(2 theta) = 2sin theta cos theta# Doesn't look helpful

#cos(2 theta) = cos^2 theta - sin^2 theta# -- Hmmmm, probably not but there are other forms for this one

#cos(2 theta) = 1 - 2sin^2 theta#. Compare this to what we started with.

If I factored out a 5, I'd have:

#5-10sin^2x = 5(1-2sin^2x)# and that's just

#5cos(2x)#

So there it is.

#5-10sin^2x = 5cos(2x)#