What is #cos^4theta+sin^3theta# in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-07-08T08:26:30

#=1/8(3+4cos2theta+2cos4 theta+6sintheta-2sin3theta)#

We will use the identities

#color(red)(>2cos^2theta=(1+cos2theta))#

#color(blue)(>4sin^3theta=3sintheta-sin3theta)#

The given expression

#=cos^4theta+sin^3theta#

#=1/4(4cos^4theta+4sin^3theta)#

#=1/4((2cos^2theta)^2+3sintheta-sin3theta)#

#=1/4((1+cos2theta)^2+3sintheta-sin3theta)#

#=1/4(1+2cos2theta+cos^2 2theta+3sintheta-sin3theta)#

#=1/4(1+2cos2theta+1/2(1+cos4 theta)+3sintheta-sin3theta)#

#=1/4(3/2+2cos2theta+cos4 theta+3sintheta-sin3theta)#

#=1/8(3+4cos2theta+2cos4 theta+6sintheta-2sin3theta)#