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NCERT Class XI Mathematics - Trigonometric Functions - Solutions

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Question : 52 of 61
Marks: +1, -0
2 cos π13\frac{\pi}{13} cos 9π13\frac{9\pi}{13} + cos 3π13\frac{3\pi}{13} + cos 5π13\frac{5\pi}{13} = 0
Solution:  
We have,
L.H.S. = 2 cos π13\frac{\pi}{13} cos 9π13\frac{9\pi}{13} + cos 3π13\frac{3\pi}{13} + cos 5π13\frac{5\pi}{13}
= cos (9π13+π13)\left(\frac{9\pi}{13}+\frac{\pi}{13}\right) + cos (9π13π13)\left(\frac{9\pi}{13}-\frac{\pi}{13}\right) + cos 3π13\frac{3\pi}{13} + cos 5π13\frac{5\pi}{13}
= cos 10π13\frac{10\pi}{13} + cos 8π13\frac{8\pi}{13} + cos 3π13\frac{3\pi}{13} + cos 5π13\frac{5\pi}{13}
= cos (13π3π13)\left(\frac{13\pi-3\pi}{13}\right) + cos (13π5π13)\left(\frac{13\pi-5\pi}{13}\right) + cos 3π13\frac{3\pi}{13} + cos 5π13\frac{5\pi}{13}
= - cos 3π13\frac{3\pi}{13} - cos (5π13)\left(\frac{5\pi}{13}\right) + cos 3π13\frac{3\pi}{13} + cos 5π13\frac{5\pi}{13}
Since cos (π - θ) = - cos θ
= 0 = R.H.S.
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