Question

Proved that -----
cos20°cos40°cos60°cos80° = 1/16
​

Answer 1

cos20°cos40°cos60°cos80°

=(1/2)(2cos20°cos40°)(1/2)cos80° [∵,cos60°=1/2]

= (1/4)[cos(20°+40°)+cos(20°-40°)]cos80°

= (1/4)(cos60°+cos20°)cos80°

= (1/4)(cos60°cos80°+cos20°cos80°)

= (1/4)(1/2)cos80°+(1/4)cos20°cos80°

= (1/8)cos80°+(1/4)(1/2)(2cos20°cos80°)

= (1/8)cos80°+(1/8)[cos(20°+80°)+cos(20°-80°)]

= (1/8)cos80°+(1/8)(cos100°+cos60°)

= (1/8)cos80°+(1/8)cos100°+(1/8)cos60°

= (1/8)(cos80°+cos100°)+(1/8)×(1/2)

= (1/8)[{2cos(80°+100°)/2}{cos(80°-100°)/2}]+(1/16)

= (1/8)(2cos90°cos10°)+(1/16)

= 0+(1/16) [cos90°=0]

= 1/16 (proved)

Hope it helps...

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Answer 2

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