How do you rewrite #cos 3theta# in terms of only #costheta# and #sintheta#?

1 Answer
Truong-Son N.
May 24, 2016

Since #cos3theta# has an odd coefficient, the double angle formula doesn't really work at first; fortunately, it derives from the additive angle formulas, which we can use:

#\mathbf(sin(upmv) = sinucosv pm cosusinv)#
#\mathbf(cos(upmv) = cosucosv ∓ sinusinv)#

Thus:

#cos3theta#

#= cos(theta+2theta)#

#= costhetacolor(red)(cos2theta) - sinthetacolor(red)(sin2theta)#

Next, we still have #cos2theta# and #sin2theta# present.

So, we have to rewrite #cos2theta# and #sin2theta# as follows, using the "double angle" formula (which is really the additive angle formula for #color(green)(u = v)#):

#color(green)(cos(theta+theta)) = costhetacostheta - sinthetasintheta#

#= color(green)(cos^2theta - sin^2theta)#

#color(green)(sin(theta+theta)) = sinthetacostheta + costhetasintheta#

#= color(green)(2sinthetacostheta)#

Thus, we end up with:

#color(blue)(cos3theta)#

#= costheta(color(green)(cos^2theta - sin^2theta)) - sintheta(color(green)(2sinthetacostheta))#

#= cos^3theta - sin^2thetacostheta - 2sin^2thetacostheta#

#= color(blue)(cos^3theta - 3sin^2thetacostheta)#

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