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

To find the maximum value of \(\sin(x + \pi/6) + \cos(x + \pi/6)\), we use trigonometric identities and properties of sine and cosine functions.

1. We start with the expression \(\sin(x + \pi/6) + \cos(x + \pi/6)\).
2. Using the angle addition formulas, we can rewrite these:
   • \(\sin(x + \pi/6) = \sin x \cdot \cos \pi/6 + \cos x \cdot \sin \pi/6\)
   • \(\cos(x + \pi/6) = \cos x \cdot \cos \pi/6 - \sin x \cdot \sin \pi/6\)
3. Substitute the values of \(\cos \pi/6 = \sqrt{3}/2\) and \(\sin \pi/6 = 1/2\): \(\sin(x + \pi/6) + \cos(x + \pi/6) = \left( \sin x \cdot \frac{\sqrt{3}}{2} + \cos x \cdot \frac{1}{2} \right) + \left( \cos x \cdot \frac{\sqrt{3}}{2} - \sin x \cdot \frac{1}{2} \right)\)
4. Simplify the expression: \(\frac{\sqrt{3}}{2} (\sin x + \cos x) + \frac{1}{2} (\cos x - \sin x)\)
5. Combine like terms: \(\frac{\sqrt{3} + 1}{2} \cos x + \frac{\sqrt{3} - 1}{2} \sin x\)
6. Recognize the form \(a \cos x + b \sin x\), where the maximum value is \(\sqrt{a^2 + b^2}\). Here, \(a = \frac{\sqrt{3} + 1}{2}\) and \(b = \frac{\sqrt{3} - 1}{2}\).
7. Calculate the maximum: \(\sqrt{\left(\frac{\sqrt{3} + 1}{2}\right)^2 + \left(\frac{\sqrt{3} - 1}{2}\right)^2}\)
8. Compute:
   • \(\left(\frac{\sqrt{3} + 1}{2}\right)^2 = \frac{(3 + 2\sqrt{3} + 1)}{4} = \frac{4 + 2\sqrt{3}}{4}\)
   • \(\left(\frac{\sqrt{3} - 1}{2}\right)^2 = \frac{(3 - 2\sqrt{3} + 1)}{4} = \frac{4 - 2\sqrt{3}}{4}\)
9. The sum is \(\frac{4 + 2\sqrt{3} + 4 - 2\sqrt{3}}{4} = \frac{8}{4} = 2\).
10. Thus, the maximum value is \(2\) and it is achieved when \(\tan x = \frac{b}{a}\), resulting in \(x = \pi/12\).

Therefore, the maximum value of \(\sin(x + \pi/6) + \cos(x + \pi/6)\) is attained at \(x = \pi/12\), corresponding to the correct answer.