Question

Consider the function $$ f(\theta)=\frac{1}{2}(a+i b) e^{i \theta}+\frac{1}{2}(a-i b) e^{-i \theta} $$ where \(a\) and \(b\) are real numbers. Show that \(f\) can be written in the form $$ f(\theta)=C_{1} \cos \theta+C_{2} \sin \theta $$ and determine the values of \(C_{1}\) and \(C_{2}\).

Short answer

Question: Rewrite the given function in terms of sine and cosine functions: \(f(\theta) = \frac{1}{2}(a + ib)e^{i\theta} + \frac{1}{2}(a - ib)e^{-i\theta}\). Answer: The given function can be written as \(f(\theta) = C_{1}\cos\theta + C_{2}\sin\theta\) with \(C_1=a\) and \(C_2=b\).

Step-by-step solution

Recall the Euler's formula

Euler's formula states that for any real number \(\theta\), $$ e^{i\theta} = \cos(\theta) + i\sin(\theta) $$ We'll use this formula to rewrite the given function in terms of sine and cosine functions.

Apply Euler's formula to the given function

Using the Euler's formula, we can rewrite the given function \(f(\theta)\) as $$ f(\theta) = \frac{1}{2}(a+ib)(\cos\theta +i\sin\theta) + \frac{1}{2}(a-ib)(\cos\theta - i\sin\theta) $$

Multiply the terms

Expanding the expression, we get $$ f(\theta) = \frac{1}{2}(a\cos\theta - b\sin\theta + i(a\sin\theta + b\cos\theta)) + \frac{1}{2}(a\cos\theta + b\sin\theta + i(a\sin\theta - b\cos\theta)) $$

Combine like terms

Adding the two terms, we get $$ f(\theta) = \frac{1}{2}(2a\cos\theta) + i\frac{1}{2}(2a\sin\theta) + \frac{1}{2}(2b\sin\theta) - i\frac{1}{2}(2b\cos\theta) $$ Simplifying, we obtain $$ f(\theta) = (a\cos\theta + b\sin\theta) + i(a\sin\theta - b\cos\theta) $$

Separate real and imaginary parts

Now, we can compare the real and imaginary parts of the function to rewrite it in terms of \(C_{1}\cos\theta + C_{2}\sin\theta\). The real part is $$ C_1 \cos\theta = a\cos\theta + b\sin\theta $$ and the imaginary part is $$ C_2 \sin\theta = a\sin\theta - b\cos\theta $$

Determine the values of \(C_{1}\) and \(C_{2}\)

Comparing the coefficients, we get the values of the constants \(C_{1}\) and \(C_{2}\) as $$ C_{1} = a, \quad C_{2} = b $$ Thus, we have successfully shown that the given function can be written as \(f(\theta) = C_{1}\cos\theta + C_{2}\sin\theta\) with \(C_1=a\) and \(C_2=b\).