Prove that cos 20°· cos40°· cos60°. cos80° = 1/16

Trigonometric ratios help us to calculate the sides of a right angled triangle with the help of a given angle and a given side. Trigonometric angle value is a collective term for values of different ratios, such as sine, cosine, tangent, secant, cotangent, and cosecant.

Answer: It's proved that cos 20° · cos40° · cos60°. cos80° = 1/16

Let us look into the explanation to understand the problem.

Explanation:

We know that,

cos 60° = 1/2 

By substituting in the LHS = cos 20°· cos40°· cos60°. cos80° of the equation we get,

cos 20° · cos40° · (1/2) . cos80° 

⇒ (1/2) cos 20° · cos40°. cos80° 

Multiplying and dividing the expression (1/2) cos 20° · cos40°. cos80°  by 2 we get,

⇒ (1/4). (2. cos 20° · cos80°). cos40° -------------- (1)

We know that, 

2 cosa cosb= cos(a+b) + cos(a-b)

Thus, 2 cos 20°cos80° = cos(20+80)° + cos(20-80)°

Substituting back to (1) we get,

LHS = (1/4) [cos(20°+80°) + cos(20°-80°)] . cos40°

= (1/4) [cos(100°) + cos(-60°)] cos40°

= (1/4) [cos(100°) + 1/2 ] cos40°         (Since, cos(-x) = cos x thus, cos(-60°) = cos 60°)

= (1/8) cos40°+ (1/4) (cos40° . cos100°) 

Multiply and divide the expression (1/8) cos40°+ (1/4) (cos40° . cos100°)  by 2, we have

(1/8)cos 40° + (1/8)(2 cos40° cos100°) --------------- (2)

We know that,

2 cosa cosb = cos(a+b) + cos(a-b)

Thus, 2 cos40° cos100° = cos(40+100)° + cos(40-100)°

Substituting back to (2) we get,

LHS = (1/8) cos 40° + (1/8) {cos(40+100)° + cos(40-100)°}

= (1/8) cos 40° + (1/8) {cos 140° + cos(-60)°}

= (1/8) cos 40° + (1/8) {cos 140° + (1/2)}

= (1/8) cos 40° + (1/8) cos 140° + 1/16

= (1/8) (cos 40° + cos 140°) + 1/16

= (1/8) {2 cos 90° cos 50°} + 1/16  (Using the formula cos(a+b) + cos(a-b) = 2cosa cosb, Here a = 90°, b = 50°)

= 0 + 1/16    (Since, cos 90° = 0)

= 1/16 = RHS

Thus, verified. cos 20° · cos40° · cos60°. cos80° = 1/16.