How Does De Moivre's Identity Help Solve Trigonometric Equations?

[subzero0137]

Use de Moivre's identity to find real values of a and b in the equation below such that the equation is valid.

$cos^6(x)+sin^6(x)+a(cos^4(x)+sin^4(x))+b=0$

Hint: Write $cos(x)$ & $sin(x)$ in terms of $e^{ix}$ & $e^{-ix}$.

Check your values of $a$ and $b$ are valid by substituting in a value of $x$. State, with explanation, two values of $x$which would not have been sufficient checks on your values of $a$ and $b$.

I've managed to obtain the following expression:

$\frac{3}{8}cos(4x)+\frac{a}{4}cos(4x)+\frac{3a}{4}+\frac{5}{8}+b=0$, and I checked the model solution, and this is the expression they've got too. But then they simply state $a=\frac{-3}{2}$, and work out $b$ from there. But I don't understand how they got that value for $a$. Can someone explain to me what I'm missing?

 

[AlephZero]

The equation $\frac{3}{8}cos(4x)+\frac{a}{4}cos(4x)+\frac{3a}{4}+\frac{5}{8}+b=0$ has to be true for all values of $x$.

So you can get two equations from it, similar to "equating coefficients" of polynomials, as in http://en.wikipedia.org/wiki/Equating_coefficients

 

[Ray Vickson]

Use de Moivre's identity to find real values of a and b in the equation below such that the equation is valid.

$cos^6(x)+sin^6(x)+a(cos^4(x)+sin^4(x)+b=0$

Hint: Write $cos(x)$ & $sin(x)$ in terms of $e^{ix}$ & $e^{-ix}$.

Check your values of $a$ and $b$ are valid by substituting in a value of $x$. State, with explanation, two values of $x$which would not have been sufficient checks on your values of $a$ and $b$.

I've managed to obtain the following expression:

$\frac{3}{8}cos(4x)+\frac{a}{4}cos(4x)+\frac{3a}{4}+\frac{5}{8}+b=0$, and I checked the model solution, and this is the expression they've got too. But then they simply state $a=\frac{-3}{2}$, and work out $b$ from there. But I don't understand how they got that value for $a$. Can someone explain to me what I'm missing?

You have
$$\left( \frac{3}{8} + \frac{a}{4} \right) \cos(4x) + \frac{5}{8} + b \equiv 0$$
where $\equiv$ means that it is an identity that holds for all $x$ (much more than just an equation!). So, the coefficient of $\cos(4x)$ must vanish.

 

[subzero0137]

You have
$$\left( \frac{3}{8} + \frac{a}{4} \right) \cos(4x) + \frac{5}{8} + b \equiv 0$$
where $\equiv$ means that it is an identity that holds for all $x$ (much more than just an equation!). So, the coefficient of $cos(4x)$ must vanish.

Okay, so I understand what AlephZero said. I picked 2 random values of x, obtained simultaneousness equations and solved them for a and b. I got a=-3/2 and b=1/2. But I don't understand the emboldened part. Why must the coefficient of $cos(4x)$ vanish?

Also, for the last part of the question where it asks for 2 values of x which would not have been sufficient checks, do I just write down the 2 random values I used before to get my simultaneous equations?

 

[Ray Vickson]

Okay, so I understand what AlephZero said. I picked 2 random values of x, obtained simultaneousness equations and solved them for a and b. I got a=-3/2 and b=1/2. But I don't understand the emboldened part. Why must the coefficient of $cos(4x)$ vanish?

Also, for the last part of the question where it asks for 2 values of x which would not have been sufficient checks, do I just write down the 2 random values I used before to get my simultaneous equations?

The coefficient of cos(4x) must vanish because the equation must hold for ALL values of x; If the coefficient of cos(4x) did not vanish you could find infinitely many different values for x that would make the equation false.

Look at it this way: you could re-write the equation as
$$\left( \frac{3}{8} + \frac{a}{4} \right) \cos(4x) = - \frac{5}{8} - b$$
The right-hand-side is a constant (not dependent on x), so the left-hand-side must also not depend on x. The only way you can make that happen is to set the coefficient of cos(4x) to zero.

 

[subzero0137]

The coefficient of cos(4x) must vanish because the equation must hold for ALL values of x; If the coefficient of cos(4x) did not vanish you could find infinitely many different values for x that would make the equation false.

Look at it this way: you could re-write the equation as
$$\left( \frac{3}{8} + \frac{a}{4} \right) \cos(4x) = - \frac{5}{8} - b$$
The right-hand-side is a constant (not dependent on x), so the left-hand-side must also not depend on x. The only way you can make that happen is to set the coefficient of cos(4x) to zero.

Got it. Thanks