Question

A.

solve

`(1)/(n) \pi = \theta - (1)/(2)sin(2 \theta)` for
`\theta` in terms of "n"

(derivation of equation below)

b. Based on your answer in
part a, if
`\theta = arccos(1 - (a)/(r) ) = {cos}^( - 1) (1 - (a)/(r) )` or
`a = r-2cos( \theta)`

find "a" as a function of
r & n. (find f(r,n)=a).

alternately, if a+b=r, we can write
`\theta = arccos( (b)/(r) ) = {cos}^( - 1) ((b)/(r) )`
then solve for "a" in terms of r and n



show all work and reasoning.
Solve analytically if possible​

Answer 1

a) There is no algebraic method for finding θ in terms of n

b) should be a = r(1 -cos(θ))

Explanation:

Algebraic methods have been developed for solving trig functions and polynomial functions individually, but not in combination. In general, the solution is easily found numerically, but not analytically.

You would be looking for the numerical solution to ...

f(n, θ) = 0

where f(n, θ) can be ...

f(n, θ) = θ - (1/2)sin(2θ) - π/n

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The attached shows Newton's method iterative solutions for n = 3 through 6:

for n = 3, θ ≈ 1.3026628373

for n = 4, θ ≈ 1.15494073001

...