Cos 10 cos 30 cos 50 cos 70
Asked by admin @ in Math viewed by 364 People
Prove that cos 10.cos30.cos50.cos70 = 3/16
Answered by admin @
(A) By the product-to-sum formula for cosine:
cos(A)cos(B) = (1/2)[cos(A + B) + cos(A - B)].
So, we have:
cos(10°)cos(30°)cos(50°)cos(70°)
= [cos(70°)cos(10°)][cos(50°)cos(30°)]
= (1/2)(1/2)[cos(80°) + cos(60°)][cos(80°) + cos(20°)], from above
= (1/4)[cos^2(80°) + cos(80°)cos(20°) + cos(80°)cos(60°) + cos(60°)cos(20°)].
Using the above formula again:
(1/4)[cos^2(80°) + cos(80°)cos(20°) + cos(80°)cos(60°) + cos(60°)cos(20°)]
= (1/4)(1/2)[cos(160°) + cos(100°) + cos(60°) + cos(140°) + cos(20°) + cos(80°) + cos(40°) + 1]
= (1/8)[cos(160°) + cos(100°) + cos(60°) + cos(140°) + cos(20°) + cos(80°) + cos(40°) + 1]
= (1/8)[cos(160°) + cos(100°) + cos(140°) + cos(20°) + cos(80°) + cos(40°)] + 3/16.
We now want to show that:
cos(160°) + cos(100°) + cos(140°) + cos(20°) + cos(80°) + cos(40°) = 0.
To do this, re-arrange the terms as follows:
[cos(160°) + cos(20°)] + [cos(140°) + cos(40°)] + [cos(100°) + cos(80°)].
Using the sum-to-product formula:
cos(A)cos(B) = 2cos[(A + B)/2]cos[(A - B)/2],
each bracketed term equals zero (as (A + B)/2 = 90° and cos(90°) = 0), so this equals:
0 + 0 + 0 = 0, as required.
Therefore, cos(10°)cos(30°)cos(50°)cos(70°) = (1/8)(0) + 3/16 = 3/16.
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