سوال ۳۱ کنکور ارشد علوم کامپیوتر ۱۳۹۷

We must simplify the ratio of two finite trigonometric sums with the same indices $1,2,\dots,6$: $\frac{\sin x+\sin 2x+\cdots+\sin 6x}{\cos x+\cos 2x+\cdots+\cos 6x}.$

Pair terms symmetrically (sum-to-product)

Pair the first with the last, the second with the fifth, and the third with the fourth. Using $\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$, every pair shares the same half-sum $\frac{7x}{2}$: $\sin x+\sin 6x=2\sin\tfrac{7x}{2}\cos\tfrac{5x}{2},$ $\sin 2x+\sin 5x=2\sin\tfrac{7x}{2}\cos\tfrac{3x}{2},$ $\sin 3x+\sin 4x=2\sin\tfrac{7x}{2}\cos\tfrac{x}{2}.$ Adding, $\text{numerator}=2\sin\tfrac{7x}{2}\left(\cos\tfrac{5x}{2}+\cos\tfrac{3x}{2}+\cos\tfrac{x}{2}\right).$

Do the same for the cosines with $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$: $\text{denominator}=2\cos\tfrac{7x}{2}\left(\cos\tfrac{5x}{2}+\cos\tfrac{3x}{2}+\cos\tfrac{x}{2}\right).$

Cancel the common factor

The bracketed sum is identical, so it cancels: $\frac{2\sin\frac{7x}{2}\,(\cdots)}{2\cos\frac{7x}{2}\,(\cdots)}=\frac{\sin\frac{7x}{2}}{\cos\frac{7x}{2}}=\tan\frac{7x}{2}.$

Closed-form check

With the Dirichlet-kernel sums $\sum_{k=1}^{n}\sin kx=\dfrac{\sin\frac{nx}{2}\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$ and $\sum_{k=1}^{n}\cos kx=\dfrac{\sin\frac{nx}{2}\cos\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$, for $n=6$ the ratio is $\dfrac{\sin\frac{7x}{2}}{\cos\frac{7x}{2}}=\tan\frac{7x}{2}$.

The answer is $\tan\dfrac{7}{2}x$ — option 4.

Exam trick: For $\dfrac{\sum_{k=1}^{n}\sin kx}{\sum_{k=1}^{n}\cos kx}$ the result is always $\tan\dfrac{(n+1)x}{2}$. Here $n=6$, so $\tan\frac{7x}{2}$ — no need to evaluate the bulky cosine sum at all.