Question

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)

Answer 2

DEAR STUDENT,

Kindly check the attached papers for a detailed solution along with step by step elaboration to reach our final solution that is our final desired or required answer to this query that is completely proved.

$\boxed{\bf{\underline{L.H.S. = R.H.S.}}}$

Which is the required proof or solution process for this type of query.

Hope this helps you and solves your doubts for applying trigonometric identifies to reach a proper solution or a proof!!!!!