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

1 Answer
Jun 13, 2016

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

Explanation:

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)