Question -
Answer -
Let us consider LHS:
cos π/15 cos 2π/15 cos 4π/15 cos 7π/15
Let us multiply and divide by 2 sin π/15, we get,
= [2 sin π/15 cos π/15] cos 2π/15 cos 4π/15 cos 7π/15]/ 2 sin π/15
We know, 2sin A cos A = sin 2A
Then the above equation becomes,
= [(sin 2π/15) cos 2π/15 cos 4π/15 cos 7π/15] / 2 sinπ/15
Now, multiply and divide by 2 we get,
= [(2 sin 2π/15 cos 2π/15) cos 4π/15 cos 7π/15] / 2 ×2 sin π/15
We know, 2sin A cos A = sin 2A
Then the above equation becomes,
= [(sin 4π/15) cos 4π/15 cos 7π/15] / 4 sin π/15
Now, multiply and divide by 2 we get,
= [(2 sin 4π/15 cos 4π/15) cos 7π/15] / 2 × 4 sin π/15
We know, 2sin A cos A = sin 2A
Then the above equation becomes,
= [(sin 8π/15) cos 7π/15] / 8 sin π/15
Now, multiply and divide by 2 we get,
= [2 sin 8π/15 cos 7π/15] / 2 × 8 sin π/15
We know, 2sin A cos B = sin (A+B) + sin (A–B)
Then the above equation becomes,
= [sin (8π/15 + 7π/15) + sin (8π/15 – 7π/15)] / 16 sinπ/15
= [sin (π) + sin (π/15)] / 16 sin π/15
= [0 + sin (π/15)] / 16 sin π/15
= sin (π/15) / 16 sin π/15
= 1/16
= RHS
Hence proved.