Question

prove that cos 10.cos30.cos50.cos70 = 3/16

Answer 1

(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.

Answer 2

$\sqrt{3}/2$ (cos10.cos50.cos70)
=$\sqrt{3}/4$ (2cos10.cos50).cos70
=$\sqrt{3}/4$ (cos60+cos40).cos70
=$\sqrt{3}/8$ (cos70+ 2cos40.cos70)
=$\sqrt{3}/8$ (cos70+ cos110+cos30)
=$\sqrt{3}/8$ (cos70 -cos70 +$\sqrt{3}/2$)
=3/16