Question

1. Prove that sin20° sin40° sin 80º =
V3/8

2. Prove that cos20° cos40° cos80°
8
3. Evaluate cos 20° + cos 100° + cos140°​

Answer 1

Answer:

Step-by-step explanation:

1. Sin20°Sin40°Sin80°

= Sin(60°-20°/2)Sin(60°+20°/2)*Sin80°

//We know that CosA - CosB = - 2Sin(A+B/2)(SinA-B/2)

=> -1/2(Cos60° - Cos20) * Sin80°

=> 1/2(Cos20° - Cos60°)Sin80°

=> 1/2(Sin80°Cos20° - Cos60°Sin80°)

=> 1/4(2Sin80°Cos20° - 2Cos60°Sin80°)

//We know that SinA + SinB = 2Sin(A+B/2)Cos(A-B/2)

=> 1/4[2Sin(100°+60°/2)Cos(100°- 60°/2) - 2Sin(140°+20°/2)Cos(140°- 20°/2)  

=> 1/4(Sin100° + Sin60° - Sin140° - Sin20°)

=> 1/4(Sin(180°-80°) + Sin60° - Sin(180°-40°) - Sin20°)

=> 1/4(Sin80° + Sin60° - Sin40° - Sin20°)  (∵ Sin(180 - Ф) = SinФ)

=> 1/4(Sin60° + Sin(60°+20°) - Sin(60° - 20°) - SIn20°)

//We know that Sin(A+B) - Sin(A-B) = 2CosASinB

=> 1/4(Sin60° + 2Cos60°Sin20° - Sin20°)

=> 1/4(Sin60° + 2*1/2*Sin20° - Sin20°)

=> 1/4(Sin60° + SIn20° - Sin20°)

= 1/4(Sin60°) = 1/4* √3/2 = √3/8.  

Hence proved.  

2. Cos20°Cos40°Cos80°

= 1/2(2Cos20°Cos40°Cos80°)

//We know that Cos(A+B) + Cos(A-B) = 2CosACosB

=> 1/2([Cos(40°+20°) + Cos(40°-20°)] Cos80°)

=>1/2(Cos60°Cos80°+Cos20°Cos80°)

=> 1/4(2Cos60°Cos80°+ 2Cos20°Cos80°)

//We know that Cos(A+B) + Cos(A-B) = 2CosACosB

=>1/4(Cos80° + Cos(20°+80°) + Cos(80°-20°))

=>1/4(Cos80° + Cos100° + Cos60°)

=>1/4(Cos80° + Cos(180°-80°) + Cos60°)

=>1/4(Cos80° - Cos80° + Cos60°)   (∵ Cos(180 - Ф) = -CosФ)

=> 1/4(Cos60°)

=>1/4*1/2

=>1/8.

Hence Proved.

3. Cos20°+ Cos100°+cos140°

//We know that CosA + CosB = 2Cos(A+B/2)Cos(A-B/2)

//We know that  Cos(180 - Ф) = -CosФ  

=> 2Cos(100​°+20°/2)Cos(100°-20°/2) + Cos(180°-40°)  

=> 2Cos60°Cos40° - Cos40°

=> Cos40° - Cos40°

=> 0