Rewriting exponential function using Euler

Question

I have the following function that I want to express using trigonometric functions:

$$f(x) = \frac{1}{2}\left(\frac{2i+2}{2i+1}e^{ix} + \frac{-2i+2}{-2i+1}e^{-ix}\right)$$

I have come so far as:

$$f(x) = \frac{1}{2}\left(\frac{2i+2}{2i+1}(\cos{x} + i\sin{x}) + \frac{-2i+2}{-2i+1}(\cos{x} - i\sin{x})\right)$$

But any simplification I do from here on just complicates things. The answer should only contain one sinus and one cosinus. I can see in the original expression that it is almost a cosinus since $\cos{x} = \frac{1}{2}(e^{ix}+e^{-ix})$ but I cannot seem to shake off those boring factors.

Answer 1

Note that $f''(x)=-f(x)$. So your function is indeed a linear combination $a\cos(x)+b\sin(x)$. Now find $a$ and $b$ by evaluating $f(0)$ and $f'(0)$.

Answer 2

First of all, note that since every term on the right hand side of the addition is the complex conjugate of the corresponding term on the LHS, your expression is $f(x)=\frac12\left(z(x)+\bar{z}(x)\right) = \mathfrak{R}(z(x))$, where $z(x) = \dfrac{2i+2}{2i+1}e^{ix}$; it's this last piece that we'll work on. To make things easier, first rationalize the denominator: $\dfrac{2i+2}{2i+1} = \dfrac{2i+2}{2i+1}\cdot\dfrac{-2i+1}{-2i+1} = \dfrac{4+2i-4i+2}{4+2i-2i+1} = \frac{1}{5}(6-2i)$. Now, use Euler's identity and multiply out again: $z(x) = \frac15(6-2i)(\cos x+i\sin x)$. I'll let you do the final multiplications and gathering of terms; once you have an expression for $z(x)$ as $a(x)+ib(x)$, you can just take the real part (as initially mentioned) to get your answer for $f(x)$.