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

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

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Answer 1 · 2016-01-15T04:09:54

$\frac{2 {sin}^{4} θ - 4 {sin}^{2} θ + sin θ sin 2 θ + 2}{2 {sin}^{2} θ}$ $(2sin^4theta-4sin^2theta+sinthetasin2theta+2)/(2sin^2theta)$

Explanation:

Write in terms of $sin θ$ $sintheta$ and $cos θ$ $costheta$ .

$= cos θ - {cos}^{2} θ + \frac{{cos}^{2} θ}{{sin}^{2} θ}$ $=costheta-cos^2theta+cos^2theta/sin^2theta$

Find a common denominator.

$= \frac{cos θ {sin}^{2} θ}{{sin}^{2} θ} - \frac{{cos}^{2} θ {sin}^{2} θ}{{sin}^{2} θ} + \frac{{cos}^{2} θ}{{sin}^{2} θ}$ $=(costhetasin^2theta)/sin^2theta-(cos^2thetasin^2theta)/sin^2theta+cos^2theta/sin^2theta$

Combine.

$= \frac{cos θ {sin}^{2} θ - {cos}^{2} θ {sin}^{2} θ + {cos}^{2} θ}{{sin}^{2} θ}$ $=(costhetasin^2theta-cos^2thetasin^2theta+cos^2theta)/sin^2theta$

The following simplification may seem unecessary, but is actually relevant. Its purpose will become clear in the following step.

$= \frac{sin θ (cos θ sin θ) - {cos}^{2} θ {sin}^{2} θ + {cos}^{2} θ}{{sin}^{2} θ}$ $=(sintheta(color(blue)(costhetasintheta))-color(green)(cos^2theta)sin^2theta+color(green)(cos^2theta))/sin^2theta$

Use the following identities:

${cos}^{2} θ = 1 - {sin}^{2} θ$ $color(green)(cos^2theta=1-sin^2theta$
$2 cos θ sin θ = sin 2 θ \Rightarrow cos θ sin θ = \frac{sin 2 θ}{2}$ $2costhetasintheta=sin2theta=>color(blue)(costhetasintheta=(sin2theta)/2$

$= \frac{sin θ (\frac{sin 2 θ}{2}) - (1 - {sin}^{2} θ) {sin}^{2} θ + (1 - {sin}^{2} θ)}{{sin}^{2} θ}$ $=(sintheta((sin2theta)/2)-(1-sin^2theta)sin^2theta+(1-sin^2theta))/sin^2theta$

$= \frac{\frac{sin θ sin 2 θ}{2} - {sin}^{2} θ + {sin}^{4} θ + 1 - {sin}^{2} θ}{{sin}^{2} θ}$ $=((sinthetasin2theta)/2-sin^2theta+sin^4theta+1-sin^2theta)/sin^2theta$

$= \frac{\frac{sin θ sin 2 θ}{2} - 2 {sin}^{2} θ + {sin}^{4} θ + 1}{{sin}^{2} θ}$ $=((sinthetasin2theta)/2-2sin^2theta+sin^4theta+1)/sin^2theta$

$= \frac{2 {sin}^{4} θ - 4 {sin}^{2} θ + sin θ sin 2 θ + 2}{2 {sin}^{2} θ}$ $=(2sin^4theta-4sin^2theta+sinthetasin2theta+2)/(2sin^2theta)$

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

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