MATH SOLVE

4 months ago

Q:
# Can someone help me in this trig question, please? thanks A person is on the outer edge of a carousel with a radius of 20 feet that is rotating counterclockwise around a point that is centered at the origin. What is the exact value of the position of the rider after the carousel rotates 5pi/12

Accepted Solution

A:

[tex]\bf \textit{the position of the rider is clearly }20cos\left( \frac{5\pi }{12} \right)~~,~~20sin\left( \frac{5\pi }{12} \right)\\\\
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\cfrac{5}{12}\implies \cfrac{2+3}{12}\implies \cfrac{2}{12}+\cfrac{3}{12}\implies \cfrac{1}{6}+\cfrac{1}{4}
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\textit{therefore then }\qquad \cfrac{5\pi }{12}\implies \cfrac{1\pi }{6}+\cfrac{1\pi }{4}\implies \cfrac{\pi }{6}+\cfrac{\pi }{4}\\\\
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[tex]\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta) \\\\ cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\ -------------------------------\\\\ cos\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=cos\left( \frac{\pi }{6}\right)cos\left(\frac{\pi }{4} \right)-sin\left( \frac{\pi }{6}\right)sin\left(\frac{\pi }{4} \right)[/tex]

[tex]\bf cos\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\cfrac{\sqrt{3}}{2}\cdot \cfrac{\sqrt{2}}{2}-\cfrac{1}{2}\cdot \cfrac{\sqrt{2}}{2}\implies \cfrac{\sqrt{6}}{4}-\cfrac{\sqrt{2}}{4}\implies \boxed{\cfrac{\sqrt{6}-\sqrt{2}}{4}} \\\\\\ sin\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=sin\left( \frac{\pi }{6}\right)cos\left( \frac{\pi }{4} \right)+cos\left( \frac{\pi }{6}\right)sin\left(\frac{\pi }{4} \right)[/tex]

[tex]\bf sin\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\cfrac{1}{2}\cdot \cfrac{\sqrt{2}}{2}+\cfrac{\sqrt{3}}{2}\cdot \cfrac{\sqrt{2}}{2}\implies \cfrac{\sqrt{2}}{4}+\cfrac{\sqrt{6}}{4}\implies \boxed{\cfrac{\sqrt{2}+\sqrt{6}}{4}}\\\\ -------------------------------\\\\ 20\left( \cfrac{\sqrt{6}-\sqrt{2}}{4} \right)\implies 5(-\sqrt{2}+\sqrt{6}) \\\\\\ 20\left( \cfrac{\sqrt{2}+\sqrt{6}}{4} \right)\implies 5(\sqrt{2}+\sqrt{6})[/tex]

[tex]\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta) \\\\ cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\ -------------------------------\\\\ cos\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=cos\left( \frac{\pi }{6}\right)cos\left(\frac{\pi }{4} \right)-sin\left( \frac{\pi }{6}\right)sin\left(\frac{\pi }{4} \right)[/tex]

[tex]\bf cos\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\cfrac{\sqrt{3}}{2}\cdot \cfrac{\sqrt{2}}{2}-\cfrac{1}{2}\cdot \cfrac{\sqrt{2}}{2}\implies \cfrac{\sqrt{6}}{4}-\cfrac{\sqrt{2}}{4}\implies \boxed{\cfrac{\sqrt{6}-\sqrt{2}}{4}} \\\\\\ sin\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=sin\left( \frac{\pi }{6}\right)cos\left( \frac{\pi }{4} \right)+cos\left( \frac{\pi }{6}\right)sin\left(\frac{\pi }{4} \right)[/tex]

[tex]\bf sin\left( \frac{\pi }{6}+\frac{\pi }{4} \right)=\cfrac{1}{2}\cdot \cfrac{\sqrt{2}}{2}+\cfrac{\sqrt{3}}{2}\cdot \cfrac{\sqrt{2}}{2}\implies \cfrac{\sqrt{2}}{4}+\cfrac{\sqrt{6}}{4}\implies \boxed{\cfrac{\sqrt{2}+\sqrt{6}}{4}}\\\\ -------------------------------\\\\ 20\left( \cfrac{\sqrt{6}-\sqrt{2}}{4} \right)\implies 5(-\sqrt{2}+\sqrt{6}) \\\\\\ 20\left( \cfrac{\sqrt{2}+\sqrt{6}}{4} \right)\implies 5(\sqrt{2}+\sqrt{6})[/tex]