Question

Prove that :
cos20° cos40° cos60° cos80° = 1/16 .

With complete solution..​

Answer 1

cos20 cos40 cos60 cos80= 1/16

l.h.s. :

cos20 cos40 1/2 cos80 (cos60 = 1/2)

multiply nd divide by 2

1/4 (2 cos20 cos40 cos80)

1/4 (cos(20+80)+ cos(20-80)) cos40 (2cosa cosb= cos(a+b) + cos(a-b))

1/4 (cos(-60)+ cos(100)) cos40

1/4(1/2 + cos100)cos40

1/8 cos40+ 1/4 (cos40 cos100)

multiplt nd divide by 2

2/2(1/8 cos40) + 1/8(2 cos40 cos100)

1/8 cos40+ 1/8 (cos140+ cos(-60)) (2cosa cosb= cos(a+b) cos(a-b))

1/8 cos40+ 1/8 cos140 + 1/16 (cos60= 1/2)

1/8(cos40+cos140) + 1/16

1/8(2 cos90 cos(-50)) + 1/16 (as above identity)

cos90= 0

1/16

= r.h.s

hence proved........

Answer 2

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)