Question

The value of $\frac{\sqrt{3}(\sin 40^{\circ}+\sin 20^{\circ})}{\cos 40^{\circ}+\cos 20^{\circ}}$ is

• A. 1
• B. $\sqrt{3}$
• C. $\sqrt{2}$
• D. $\frac{1}{2}$
• E. 0

Hint:

Logic Tip: Notice that $\frac{\sin A + \sin B}{\cos A + \cos B}$ will always simplify neatly to $\tan\left(\frac{A+B}{2}\right)$. Using this shortcut, the expression instantly becomes $\sqrt{3} \cdot \tan(30^{\circ}) = 1$.

Answer 1

The Correct Option is A

Concept:
This problem requires the sum-to-product trigonometric identities: $$\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$ $$\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$
Step 1: Apply the sum-to-product formula to the numerator.
Let $A = 40^{\circ}$ and $B = 20^{\circ}$. $$\sin 40^{\circ} + \sin 20^{\circ} = 2\sin\left(\frac{40^{\circ}+20^{\circ}}{2}\right)\cos\left(\frac{40^{\circ}-20^{\circ}}{2}\right)$$ $$= 2\sin(30^{\circ})\cos(10^{\circ})$$
Step 2: Apply the sum-to-product formula to the denominator.
Using the same $A$ and $B$: $$\cos 40^{\circ} + \cos 20^{\circ} = 2\cos\left(\frac{40^{\circ}+20^{\circ}}{2}\right)\cos\left(\frac{40^{\circ}-20^{\circ}}{2}\right)$$ $$= 2\cos(30^{\circ})\cos(10^{\circ})$$
Step 3: Substitute back into the fraction and simplify.
Original expression: $$\frac{\sqrt{3}[2\sin(30^{\circ})\cos(10^{\circ})]}{2\cos(30^{\circ})\cos(10^{\circ})}$$ The common terms $2$ and $\cos(10^{\circ})$ cancel out from the numerator and denominator: $$= \sqrt{3} \cdot \frac{\sin(30^{\circ})}{\cos(30^{\circ})}$$ $$= \sqrt{3} \cdot \tan(30^{\circ})$$
Step 4: Evaluate the standard trigonometric values.
We know that $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$. $$= \sqrt{3} \cdot \left(\frac{1}{\sqrt{3}}\right) = 1$$