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?

 

[CellCoree]

Sure, I can help you get started! Let's break down the problem into smaller parts so it's easier to understand.

a) The first part of the problem is asking you to solve the trigonometric equation A = cosX + AsinX for some angle X. This means that you need to find the value of X that satisfies this equation. To do this, you can use some trigonometric identities and properties.

First, we can rewrite the equation as A = cosX(1 + AsinX). Then, we can use the identity cos^2X + sin^2X = 1 to get A = cosX(1 + AsinX) = cosX + AsinXcosX. Next, we can use the identity sinXcosX = 1/2sin2X to get A = cosX + 1/2Asin2X. Finally, we can use the inverse trigonometric function to solve for X. So we have: X = arccos(A - 1/2Asin2X).

b) Now, the second part of the problem asks you to solve for i by using the equation you found in part a). Remember that i = √-1, so we can substitute this into the equation to get: √-1 = cosX + i(sinX). Now, we can use the Euler identity e^iX = cosX + isinX to get: √-1 = e^iX. To solve for X, we can take the natural logarithm of both sides, giving us: ln(√-1) = ln(e^iX). Using the property of logarithms, we can rewrite this as: ln(√-1) = iX. Finally, we can solve for X by dividing both sides by i, giving us: X = ln(√-1)/i.

c) The last part of the problem asks you to evaluate i^i for one value. To do this, we can use the exponent arithmetic property (a^b)^c = a^(bc) to get: i^i = (e^iπ/2)^i. Then, we can use the property of logarithms again to rewrite this as: i^i = e^(i^2π/2) = e^(-π/2). So, the value of i^i for one