4cos12°cos48°-cos72°
Answers
Answer:
To prove : 4 \cos 12^\circ \cos 48^\circ \cos 72^\circ = \cos 36^\circ4cos12
∘
cos48
∘
cos72
∘
=cos36
∘
Proof :
Taking LHS,
4 \cos 12^\circ \cos 48^\circ \cos 72^\circ4cos12
∘
cos48
∘
cos72
∘
=(2\cos 48^\circ \cos 12^\circ)\times 2\cos 72^\circ=(2cos48
∘
cos12
∘
)×2cos72
∘
Using identity, 2\cos A\cos B=\cos(A+B)+\cos(A-B)2cosAcosB=cos(A+B)+cos(A−B)
=(\cos (48+12)^\circ+\cos (48-12)^\circ)\times 2\cos 72^\circ=(cos(48+12)
∘
+cos(48−12)
∘
)×2cos72
∘
=(\cos (60)^\circ+\cos (36)^\circ)\times 2\cos 72^\circ=(cos(60)
∘
+cos(36)
∘
)×2cos72
∘
=(\frac{1}{2}+\cos (36)^\circ)\times 2\cos 72^\circ=(
2
1
+cos(36)
∘
)×2cos72
∘
=\cos 72+2\cos 72^\circ\cos (36)^\circ=cos72+2cos72
∘
cos(36)
∘
=\cos 72+\cos (72+36)^\circ+\cos (72-36)^\circ=cos72+cos(72+36)
∘
+cos(72−36)
∘
=\cos 72+\cos (108)^\circ+\cos (36)^\circ=cos72+cos(108)
∘
+cos(36)
∘
Using identity, \cos C+\cos D=2\cos\frac{(C+D)}{2}\cos\frac{(C-D)}{2}cosC+cosD=2cos
2
(C+D)
cos
2
(C−D)
=2\cos \frac{(108+72)}{2}\cos \frac{(108-72)}{2}+\cos (36)^\circ=2cos
2
(108+72)
cos
2
(108−72)
+cos(36)
∘
=2\cos \frac{180}{2}\cos \frac{36}{2}+\cos (36)^\circ=2cos
2
180
cos
2
36
+cos(36)
∘
=2\cos 90\cos 18+\cos (36)^\circ=2cos90cos18+cos(36)
∘
=2\times 0\times \cos 18+\cos (36)^\circ=2×0×cos18+cos(36)
∘
=\cos (36)^\circ=cos(36)
∘
=RHS
Hence proved.