Question

The maximum value of $\sin(x + \pi/6) + \cos(x + \pi/6)$ is attained at $x =$

• A. $\pi/2$
• B. $\pi/4$
• C. $\pi/6$
• D. $\pi/12$

Hint:

For $a\sin \theta + b\cos \theta$, if $a=1, b=1$, the maximum occurs when $\theta = 45^\circ$ or $\pi/4$. Here $\theta = (x + 30^\circ)$. So, $x + 30^\circ = 45^\circ \implies x = 15^\circ$, which is $\pi/12$. Working in degrees can often be faster for mental math than manipulating $\pi$ fractions.

Answer 1

The Correct Option is D

Step 1: Let the expression be simplified
Given, \[ f(x) = \sin\left(x + \frac{\pi}{6}\right) + \cos\left(x + \frac{\pi}{6}\right) \] Let, \[ \theta = x + \frac{\pi}{6} \] Then, \[ f(x) = \sin\theta + \cos\theta \]
Step 2: Convert into single trigonometric form
We use the identity: \[ \sin\theta + \cos\theta = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right) \] So, \[ f(x) = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right) \] Substitute $\theta = x + \frac{\pi}{6}$: \[ f(x) = \sqrt{2}\sin\left(x + \frac{\pi}{6} + \frac{\pi}{4}\right) \] \[ = \sqrt{2}\sin\left(x + \frac{5\pi}{12}\right) \]
Step 3: Find maximum condition
The maximum value of $\sin(\alpha)$ is $1$, which occurs when: \[ \alpha = \frac{\pi}{2} \] So, \[ x + \frac{5\pi}{12} = \frac{\pi}{2} \]
Step 4: Solve for $x$
\[ x = \frac{\pi}{2} - \frac{5\pi}{12} \] \[ x = \frac{6\pi}{12} - \frac{5\pi}{12} = \frac{\pi}{12} \]
Step 5: Final Answer
\[ \boxed{x = \frac{\pi}{12}} \]