Question

prove that cos20°.cos40°.cos60°.cos80°=1÷16

Answer 1

since cos 60 = 1/2

LHS=1/2 cos20cos40cos80

multiplying and dividing by 2,

=>1/4 cos80(2cos40cos20)

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

= 1/4 cos80(cos60+cos20)

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

opening the bracket:

=> 1/8 cos80 + 1/4 cos80cos20

multiplying and dividing [1/4 cos80cos20] by 2,

=> 1/8 cos80 + 1/8 (2cos80cos20)

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

=1/8 cos80 + 1/8 (cos100 + 1/2
=1/8 cos80 + 1/8 cos100 + 1/16

 since cos100 = cos (180-80) = -cos80,

=> 1/8 cos80 + 1/8 (-cos80) +1/16

= 1/8 cos80 - 1/8 cos80 + 1/16

= 1/16 = RHS

Answer 2

cos 20 × cos 40 × cos 60 × cos 80 = 1/16

L.H.S.

cos 30 ×cos 40 × 1/2× cos 80

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

hence R.H.S.
proved.