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

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Answer 1 · 2016-02-01T22:43:07

$(sin θ + cos θ) (sin θ - cos θ) + (csc θ + 1) (csc θ - 1)$ $(sintheta+costheta)(sintheta-costheta)+(csctheta+1)(csctheta-1)$

Rewrite to group the sine and cosine terms.

$= {sin}^{4} θ - {cos}^{4} θ + {cot}^{2} θ$ $=sin^4theta-cos^4theta+cot^2theta$

Simplify the first two terms as a difference of squares.

$= ({sin}^{2} θ + {cos}^{2} θ) ({sin}^{2} θ - {cos}^{2} θ) + {cot}^{2} θ$ $=(sin^2theta+cos^2theta)(sin^2theta-cos^2theta)+cot^2theta$

Note that ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$ through the Pythagorean Identity.

$= {sin}^{2} θ - {cos}^{2} θ + {cot}^{2} θ$ $=sin^2theta-cos^2theta+cot^2theta$

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

$= (sin θ + cos θ) (sin θ - cos θ) + {cot}^{2} θ$ $=(sintheta+costheta)(sintheta-costheta)+cot^2theta$

Use the identity: ${cot}^{2} θ + 1 = {csc}^{2} θ$ $cot^2theta+1=csc^2theta$ to say that ${cot}^{2} θ = {csc}^{2} θ - 1$ $cot^2theta=csc^2theta-1$ .

$= (sin θ + cos θ) (sin θ - cos θ) + {csc}^{2} θ - 1$ $=(sintheta+costheta)(sintheta-costheta)+csc^2theta-1$

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

$= (sin θ + cos θ) (sin θ - cos θ) + (csc θ + 1) (csc θ - 1)$ $=(sintheta+costheta)(sintheta-costheta)+(csctheta+1)(csctheta-1)$

How do you express sin4θ+cot2θ−cos4θsin^4theta+cot^2theta -cos^4 theta in terms of non-exponential trigonometric functions?