briscosarg19
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Use a Pythagorean identity to rewrite the equation below using only the function sin θ. Then find the value of r if θ=30°,60°, and 90°.

r=4*sinθ*cos^2θ

Please explain/show steps, thanks!

Respuesta :

[tex]\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\ -------------------------------\\\\ r=4sin(\theta )cos^2(\theta )\implies r=4sin(\theta )[1-sin^2(\theta )] \\\\\\ r=4sin(\theta )-4sin^3(\theta )\qquad \qquad \begin{cases} r=4sin(30^o)-4sin^3(30^o)\\ r=4sin(60^o)-4sin^3(60^o)\\ r=4sin(90^o)-4sin^3(90^o) \end{cases}[/tex]

[tex]\bf \begin{cases} r=4\left( \frac{1}{2} \right)-4\left( \frac{1}{2} \right)^3\\\\ r=4\left( \frac{\sqrt{3}}{2} \right)-4\left( \frac{\sqrt{3}}{2} \right)^3\\\\ r=4(1)-4(1)^3 \end{cases}\implies \begin{cases} r=2-\frac{1}{2}\\\\ r=2\sqrt{3}-\frac{3\sqrt{3}}{2}\\\\ r=4-4 \end{cases}\implies \begin{cases} r=\frac{1}{2}\\\\ r=\frac{\sqrt{3}}{2}\\\\ r=0 \end{cases}[/tex]

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