Prove that 4 cos 12° cos 48° cos 72° = cos 36°.
Answers
Answer:
Step-by-step explanation:
LHS
=4cos12°×cos48°×cos72°
=(2cos48°cos12°)×2cos72°
=[cos(48°+12°)+cos(48°-12°)]×2cos72° [2cosAcosB=cos(A+B)+cos(A-B)]
=(cos60°+cos36°)×2cos72°
=1/2×2cos72°+2cos72°cos36°
=cos72°+[cos(72°+36°)+cos(72°-36°)]
=cos72°+cos108°+cos36°
=2cos(108°+72°)/2cos(108°-72°)/2+cos36° [cosC+cosD=2cos(C+D)/2cos(C-D)/2]
=2cos(180°/2)cos(36°/2)+cos36°
=2cos90°cos18°+cos36°
=2×0×cos18°+cos36°
=cos36°=RHS (Proved)
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L.H.S = R.H.S.
Step-by-step explanation:
In the question
Given 4 cos 12° cos 48° cos 72° = cos 36°
Now L.H.S. = 4 cos 12° cos 48° cos 72°
The given expression can be written as:
2 cos 72° [2 cos 12° cos 48°]
= 2 cos 72° [cos(48° + 12°) + cos(48° - 12°)]
{ Since we know that, 2 cos A cos B = cos(A + B) + cos(A - B)}
= 2 cos 72° [cos 60° + cos 36°]
= 2 cos 72° [1/2 + cos 36°] {∵ cos 60° = 1/2}
= cos 72° + 2 cos 72° cos 36°
= cos 72° [cos(72° + 36°) + cos(72° - 36°)]
= cos 72° + cos 108° + cos 36°
= 2 cos(108° + 72°)/2 cos(108° - 72°)/2 + cos 36°
{since we know that, cos C + cos D = 2 cos(C + D)/2 cos(C - D)/2}
= 2 cos 180°/2 cos 36°/2 + cos 36°
= 2 cos 90° cos 18° + cos 36°
= 0 + cos 36° {∵ cos 90° = 0}
= cos 36°