Question

$2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)$, then $\tan \theta=$

• A. $-\frac{1}{\sqrt{3}}$
• B. $-\sqrt{3}$
• C. $\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Hint:

Alternatively, use the co-function identity: $\cos\left(\theta - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{2} - (\theta - \frac{\pi}{6})\right) = \sin\left(\frac{2\pi}{3} - \theta\right)$. You can then expand this to avoid dealing with both sines and cosines initially!

Answer 1

The Correct Option is B

Step 1: Open out both sides.
Use the sum and difference formulas on $2\sin(\theta + \tfrac{\pi}{3}) = \cos(\theta - \tfrac{\pi}{6})$.
With $\cos\tfrac{\pi}{3} = \tfrac12$, $\sin\tfrac{\pi}{3} = \tfrac{\sqrt3}{2}$, $\cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}$, $\sin\tfrac{\pi}{6} = \tfrac12$.

Step 2: Write the expanded equation.
$$\sin\theta + \sqrt3\cos\theta = \tfrac{\sqrt3}{2}\cos\theta + \tfrac12\sin\theta$$

Step 3: Gather like terms.
$$\tfrac12\sin\theta = -\tfrac{\sqrt3}{2}\cos\theta \Rightarrow \sin\theta = -\sqrt3\cos\theta$$
Dividing by $\cos\theta$ gives $\tan\theta = -\sqrt3$.
\[ \boxed{-\sqrt3} \]