MHB Solve the trigonometric equation?

[arroww]

a) Solve the trigonometric equation: A = cosX + AsinX, for some angle X.

b) The imaginary number i is equal to √-1. Use part a) to solve i = cosX + isinX.

You will have used degrees to answer parts a) and b). However, the mathematics below requires the use of radians. Convert degrees to radians by multiplying the degree measurement by π/180.

You will now find a value of i^i.

This will require the euler identity e^iX = cosX + isinX, part b) and exponent arithmetic.

c) Evaluate, for one value, i^i.

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So I honestly have no idea how to start this off haha. Could someone help out? Obviously you don't need to solve the whole thing - i just need to know how to start it off.Thanks! (:

 

[Ackbach]

$a \sin(x)+b \cos(x)=c \sin(x+\varphi),$ where $c^{2}=a^{2}+b^{2}$ and $\varphi=\text{atan}2(b,a)$. That last is not a typo - it's the atan2 function, available in many programming languages. It essentially resolves the ambiguity inherent in the arctangent function.

 

[Prove It]

a) Solve the trigonometric equation: A = cosX + AsinX, for some angle X.

b) The imaginary number i is equal to √-1. Use part a) to solve i = cosX + isinX.

You will have used degrees to answer parts a) and b). However, the mathematics below requires the use of radians. Convert degrees to radians by multiplying the degree measurement by π/180.

You will now find a value of i^i.

This will require the euler identity e^iX = cosX + isinX, part b) and exponent arithmetic.

c) Evaluate, for one value, i^i.

----

So I honestly have no idea how to start this off haha. Could someone help out? Obviously you don't need to solve the whole thing - i just need to know how to start it off.Thanks! (:

I would proceed like this:

$\displaystyle \begin{align*} A &= \cos{(x)} + A\sin{(x)} \\ A - A\sin{(x)} &= \cos{(x)} \\ A \left[ 1 - \sin{(x)} \right] &= \cos{(x)} \\ \left\{ A \left[ 1 - \sin{(x)} \right] \right\} ^2 &= \cos^2{(x)} \\ A^2 \left[ 1 - 2\sin{(x)} + \sin^2{(x)} \right] &= 1 - \sin^2{(x)} \\ A - 2A\sin{(x)} + \sin^2{(x)} &= 1 - \sin^2{(x)} \\ 2\sin^2{(x)} - 2A\sin{(x)} &= 1 - A \\ \sin^2{(x)} - A\sin{(x)} &= \frac{1 - A}{2} \\ \sin^2{(x)} - A\sin{(x)} + \left( -\frac{A}{2} \right) ^2 &= \frac{1 - A}{2} + \left( -\frac{A}{2} \right) ^2 \\ \left[ \sin{(x)} - \frac{A}{2} \right] ^2 &= \frac{A^2 - 2A + 2}{4} \\ \sin{(x)} - \frac{A}{2} &= \pm \frac{\sqrt{ A^2 - 2A + 2}}{2} \\ \sin{(x)} &= \frac{A \pm \sqrt{A^2 - 2A + 2}}{2} \\ x &= \arcsin{ \left( \frac{A \pm \sqrt{A^2 - 2A + 2}}{2} \right) } \end{align*}$

as well as all other quadrant solutions and all other possibilities when the period of $\displaystyle \begin{align*} 2\pi \end{align*}$ is added or subtracted.

Just remember that not all solutions might be valid, as squaring to solve an equation might bring in extraneous solutions. You need to check which are possible.

 

[Opalg]

For a), my solution starts like Prove It's, but then diverges from it.
$$A = \cos x + A\sin x,$$ $$A(1-\sin x) = \cos x,$$ $$A^2(1-\sin x)^2 = \cos^2x = 1 - \sin^2x = (1-\sin x)(1+\sin x),$$ $$A^2(1-\sin x) = 1+\sin x,\quad(*)$$ $$(A^2+1)\sin x = A^2-1,$$ $$\sin x = \frac{A^2-1}{A^2+1},$$ $$x = \arcsin\Bigl(\frac{A^2-1}{A^2+1}\Bigr)\quad(**).$$

For b), the obvious thing to do is to put $A=i$ in a). But then we hit a serious roadblock, because if $A=i$ then $A^2 = -1$ and $A^2+1=0.$ So the denominator of the fraction in (**) is zero. What has gone wrong here?? To see what to do about this, notice that to get (*) from the previous line, I canceled a factor $(1-\sin x)$ from both sides of the equation. When might that step not be justified?