Question

cos(π/10-A)*cos(π/10+A) + cos(2π/5-A)*cos(2π/5+A) = cos2A
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Answer 1

Step-by-step explanation:

$=> \small{cos\big(\frac{\pi}{10}-\small{A}\big)*cos\big(\frac{\pi}{10}+\small{A}\big) + cos\big(\frac{2\pi}{5}-\small{A}\big)*cos\big(\frac{2\pi}{5}+\small{A}\big) }\\\\\\=> \small{cos\big(\frac{\pi}{10}-\small{A}\big)*cos\big(\frac{\pi}{10}+\small{A}\big) + cos\big(\frac{\pi}{2}-\big(\frac{2\pi}{5}-\small{A}\big)\big)*cos\big(\frac{\pi}{2}-\big(\frac{2\pi}{5}+\small{A}\big)\big) }$

$=> \small{cos\big(\frac{\pi}{10}-\small{A}\big)*cos\big(\frac{\pi}{10}+A\big)+sin(\frac{\pi}{10}+A\big)*sin\big(\frac{\pi}{10}-A \big)}\\\\\\=> cos\big(\frac{\pi}{10}-A-\frac{\pi}{10}-A\big)\\\\\\=>cos(-2A)\\\\=> cos(2A)$