How do you express #sin^4theta+cot^2theta -cos^4 theta# in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-02-01T22:43:07

#(sintheta+costheta)(sintheta-costheta)+(csctheta+1)(csctheta-1)#

Rewrite to group the sine and cosine terms.

#=sin^4theta-cos^4theta+cot^2theta#

Simplify the first two terms as a difference of squares.

#=(sin^2theta+cos^2theta)(sin^2theta-cos^2theta)+cot^2theta#

Note that #sin^2theta+cos^2theta=1# through the Pythagorean Identity.

#=sin^2theta-cos^2theta+cot^2theta#

The first two terms can again be factored as a difference of squares.

#=(sintheta+costheta)(sintheta-costheta)+cot^2theta#

Use the identity: #cot^2theta+1=csc^2theta# to say that #cot^2theta=csc^2theta-1#.

#=(sintheta+costheta)(sintheta-costheta)+csc^2theta-1#

Again, the last two terms can be factored as a difference of squares.

#=(sintheta+costheta)(sintheta-costheta)+(csctheta+1)(csctheta-1)#