Question

cos^2 21°+ cos^2 39°-cos21°cos39°

Answer 1

Cos²21°+Cos²39°-Cos21°Cos39°
=1/2(2cos²21°+2cos²39°-2cos21°cos39°)
=1/2[1+cos42°+1+cos78°-{cos(21°+39°)+cos(21°-39°)}] 
 [∵, 2cos²Ф=1+cos2Ф and 2cosAcosB=cos(A+B)+cos(A-B)]
=1/2[2+cos42°+cos78°-cos60°-cos18°]
=1/2[(2-1/2)+cos42°-cos18°+cos78°] ; [∵, cos60°=1/2]
=1/2[3/2+{2sin(42°+18°)/2sin(18°-42°)}+cos78°]
[∵, cosC-cosD=2sin(C+D)/2sin(D-C)/2]
=1/2[3/2+2sin30°sin(-12°)+cos(90°-12°)]
=1/2[3/2+2.1/2.(-sin12°)+sin12°] ; [∵, sin30°=1/2 and cos(90°-Ф)=sinФ]
=1/2[3/2-sin12°+sin12°]
=1/2×3/2
=3/4

Answer 2

Answer:

right answer is in the attachment