Auxiliary Angle Method Given , find .

Trigonometry Solution – Problem 9: find

Auxiliary Angle Method

Given $2\sin\alpha + 1 = 2\sqrt{3}\cos\alpha$ , find $\sin\left(2\alpha + \frac{\pi}{6}\right)$ .

From $2\sin\alpha + 1 = 2\sqrt{3}\cos\alpha$ :

$4\left(\sin\alpha + \frac{\sqrt{3}}{2}\cos\alpha\right) = 1$

$\sin\left(\alpha + \frac{\pi}{3}\right) = \frac{1}{4}$

Therefore:

$\sin\left(2\alpha + \frac{\pi}{6}\right) = \sin\left[2\left(\alpha + \frac{\pi}{3}\right) - \frac{\pi}{2}\right]$

$= -\cos\left[2\left(\alpha + \frac{\pi}{3}\right)\right]$

$= -\left(1 - 2\sin^2\left(\alpha + \frac{\pi}{3}\right)\right)$

$= 2 \cdot \left(\frac{1}{4}\right)^2 - 1 = \frac{1}{8} - 1 = \frac{7}{8}$

$\frac{7}{8}$

Correct identity setup (2 pts): choose an appropriate sum/difference, double-angle, or auxiliary-angle idea and set up the key equation(s).
Correct algebra / trig simplification (2 pts): transform expressions without sign mistakes.
Solve for target quantity (2 pts): isolate the requested value and handle any constraints if needed.
Final answer (1 pt): clearly state the result in the required form.