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.

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

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