Trigonometric Sum PYQ | JEE Main 2025 | Fast Identity Trick

 

❓ Question

Evaluate

$$\cos\frac{\pi}{11} +\cos\frac{3\pi}{11} +\cos\frac{5\pi}{11} +\cos\frac{7\pi}{11} +\cos\frac{9\pi}{11}.$$

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✍️ Solution

Multiply the expression by

$$2\sin\frac{\pi}{11}$$

and use the identity

$$2\sin A\cos B=\sin(A+B)+\sin(A-B).$$

Thus,

$$2S\sin\frac{\pi}{11} = \sum_{k=1,3,5,7,9} \left[ \sin\frac{(k+1)\pi}{11} + \sin\frac{(1-k)\pi}{11} \right].$$

Expanding,

$$\begin{aligned} 2S\sin\frac{\pi}{11} &= \sin\frac{2\pi}{11}+\sin0 +\sin\frac{4\pi}{11}-\sin\frac{2\pi}{11}\\ &\quad +\sin\frac{6\pi}{11}-\sin\frac{4\pi}{11} +\sin\frac{8\pi}{11}-\sin\frac{6\pi}{11}\\ &\quad +\sin\frac{10\pi}{11}-\sin\frac{8\pi}{11}. \end{aligned}$$

All intermediate terms cancel, leaving

$$2S\sin\frac{\pi}{11} = \sin0+\sin\frac{10\pi}{11}.$$

Since

$$\sin0=0,$$

and

$$\sin\frac{10\pi}{11} = \sin\left(\pi-\frac{\pi}{11}\right) = \sin\frac{\pi}{11},$$

we get

$$2S\sin\frac{\pi}{11} = \sin\frac{\pi}{11}.$$

Hence,

$$2S=1$$

and therefore

$$\boxed{S=\frac12.}$$

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✅ Final Answer

$$\boxed{\frac12}$$

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