Mathematics, 23.06.2019 07:00 camillexv2668

A. solve $\frac{1}{n} \pi = \theta - \frac{1}{2}sin(2 \theta)$ for $\theta$ in terms of "n"(derivation of equation below)b. based on your answer inpart a, if $\theta = arccos(1 - \frac{a}{r} ) = {cos}^{ - 1} (1 - \frac{a}{r} )$ or $a = r-2cos( \theta)$ find "a" as a function ofr & n. (find f(r, n)=a).alternately, if a+b=r, we can write $\theta = arccos( \frac{b}{r} ) = {cos}^{ - 1} (\frac{b}{r} )$ then solve for "a" in terms of r and nshow all work and reasoning. solve analytically if possible

$A.solve [tex]\frac{1}{n} \pi = \theta - \frac{1}{2}sin(2 \theta)[/tex] for [tex] \theta[/tex] in ter$

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A. solve [tex]\frac{1}{n} \pi = \theta - \frac{1}{2}sin(2 \theta)[/tex] for [tex] \theta[/tex] in te...