What is cos5θ6sin3θ in terms of non-exponential trigonometric functions?

1 Answer
Dec 18, 2015

Use these trig identities to transform cos5x and sin3x
cos2x=2cos2x1=12sin2x
sin2x+cos2x=1

(1) --> cos5x=cosx(cos2x)(cos2x)=
=cosx(1sinx)(1+sinx)(1+cos2x2)
(2) --> 6sin3x=6sinx(sin2x)=6sinx(1cosx)(1+cosx).
f(x) = (1) + (2)
f(x)=(12)cosx(1sinx)(1+sinx)(1+cos2x)6sinx(1cosx)(1+cosx)