How do you express #sin^4theta-cos^3(pi/2-theta ) # in terms of non-exponential trigonometric functions?

Question

1891 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-13T08:40:06

#=1/4(1-2cos2theta+1/2(1+cos4theta) -3sintheta+sin3theta)#

Given
#=sin^4theta-cos^3(pi/2-theta)#

#=sin^4theta-sin^3theta#

Now we know
#sin^2theta =1/2(1-cos2theta)#
and
#sin^3theta=1/4(3sintheta-sin3theta)#

Inserting these values the given expression becomes

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

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

#=1/4(1-2cos2theta+1/2(1+cos4theta) -3sintheta+sin3theta)#