Question

cos 12° + cos 60° + cos 84° = cos 24° + cos 48°.
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Answer 1

Hi friend,

cos12°+cos60° + cos 84° = cos24°+cos48°

=cos24 - cos 84 + cos 48 - cos 12=cos60

= 2sin 54°× sin30 + 2 sin 30 × sin (-18) = cos 60;

= sin 54 - sin 18 = cos 60

= 2 cos 36 × sin18 = cos 60

= 2×1/4 ( root 5 + 1 ) × 1/4 ( root 5 - 1 ) = 1/2

= 1/2 × ( 5 - 1 ) × 1/4 = 1/2

=1/2 = 1/2

i hope it helps

if it works you mark me as a brainlist.

Answer 2

$\green{Answer}$

cos 12 + cos 60 + cos 84 = cos 24 + cos 48

=> cos 24 - cos 84 + cos48 - cos 12 = cos 60

=> 2sin 54 * sin 30 + 2 sin 30 * sin (-18) = cos 60 ; use CosC - CosD = 2Sin[(8C+D)/2]Sin[(D-C)/2]

=> sin 54 - sin 18 = cos 60 ; sin (-θ) = - sin θ, and use sin 30 = ½

=> 2cos 36 * sin18 = cos 60 ; use SinC - SinD = 2Cos[(C+D)/2]Sin[(C-D)/2]

=> 2*¼(√5 + 1)* ¼(√5 - 1) = ½ ; use cos 36 = ¼(√5 + 1) and sin 18 = ¼(√5 - 1)

=> ½*(5 - 1)*¼ = ½

=> ½ = ½

$\green{Hence Proved}$