prove that Cos20 Cos40 Cos60 Cos80 = 1/16
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Consider the given equation
Cos 20° · Cos40° · Cos60° . Cos80° = 1/16
LHS = Cos 20° · Cos40° · Cos60° . Cos80°
We know that Cos60° = 1/2
LHS = Cos 20° · Cos40° · 1/2 . Cos80°
Multiply and divide the equation by 2
LHS = 1/4 (2. Cos 20° · Cos40° · Cos80°)
We know the formula 2 cosa cosb= cos(a+b) + cos(a-b)
LHS = 1/4 [Cos(20+80)+ Cos(20-80)] . Cos40
LHS = 1/4 [Cos(-60)+ Cos(100)] Cos40
LHS = 1/4 [1/2 + cos100] Cos40
LHS = 1/8 Cos40+ 1/4 (Cos40 . Cos100)
Multiply and divide the equation by 2
LHS = 2/2 (1/8 Cos 40) + 1/8(2 Cos40 Cos100)
We know the formula
2cosa cosb= cos(a+b) cos(a-b)
LHS = 1/8 Cos40+ 1/8 [Cos 140 + Cos (-60)]
LHS = 1/8 Cos 40+ 1/8 Cos 140 + 1/16
Since Cos 60= 1/2
LHS = 1/8 (Cos 40 + Cos 140) + 1/16
LHS = 1/8 [2 Cos 90 Cos (-50)] + 1/16
LHS = Cos 90
Cos 90 = 0
LHS = 1/16
LHS = RHS