Trig Functions — Question 3

Question 3

Consider the function $$f(x) = \cos(2x) - \sin(2x)$$

• Write $f(x)$ as a single trigonometric function of the form $A\cos(2x + \phi)$.

• Determine the amplitude and phase shift of the resulting function.

• Find the maximum and minimum values of $f(x)$.

• Determine all $x \in [0, 2\pi]$ such that $f(x) = 0$.

Question 3 - Solution

(a) Write $f(x)$ as a single cosine function:

We use the identity: $$A\cos(2x + \phi) = A\cos\phi \cos(2x) - A\sin\phi \sin(2x)$$

Match coefficients with: $$\cos(2x) - \sin(2x)$$

This gives the system: $$A\cos\phi = 1, \quad A\sin\phi = 1$$

Square and add: $$A^2(\cos^2\phi + \sin^2\phi) = 1^2 + 1^2 = 2 \Rightarrow A = \sqrt{2}$$

Now: $$\cos\phi = \frac{1}{\sqrt{2}}, \quad \sin\phi = \frac{1}{\sqrt{2}} \Rightarrow \phi = \frac{\pi}{4}$$

So: $$f(x) = \sqrt{2}\cos\left(2x + \frac{\pi}{4}\right)$$

(b) Amplitude and Phase Shift:

From the rewritten form:

• Amplitude: $\boxed{\sqrt{2}}$

• Phase shift: Solve $2x + \frac{\pi}{4} = 0 \Rightarrow x = -\frac{\pi}{8}$

So the graph is shifted $\boxed{\frac{\pi}{8}}$ units to the left.

(c) Maximum and Minimum Values:

Since cosine ranges between $-1$ and $1$: $$\text{Maximum} = \sqrt{2}, \quad \text{Minimum} = -\sqrt{2}$$

(d) Solve $f(x) = 0$ on $[0, 2\pi]$:

$$\cos(2x) - \sin(2x) = 0 \Rightarrow \cos(2x) = \sin(2x) \Rightarrow \tan(2x) = 1$$

$$2x = \frac{\pi}{4} + n\pi \Rightarrow x = \frac{\pi}{8} + \frac{n\pi}{2}$$

Find values in $[0, 2\pi]$:

$$x = \frac{\pi}{8},\ \frac{5\pi}{8},\ \frac{9\pi}{8},\ \frac{13\pi}{8}$$

Answer: $$\boxed{x = \frac{\pi}{8},\ \frac{5\pi}{8},\ \frac{9\pi}{8},\ \frac{13\pi}{8}}$$