Question

4 cos 12 cos 48 cos 72 = cos 36

Answer 1

4cos12°cos48°cos72°
=2(2cos12°cos48°)cos72°
=2[cos(12°+48°)+cos(12°-48°)]cos72°
=2(cos60°+cos36°)cos72°
=2cos60°cos72°+2cos36°cos72°
=2×1/2×cos72°+[cos(36°+72°)+cos(36°-72°)]
=cos72°+cos108°+cos36°
=2cos(72°+108°)/2cos(72°-108°)/2+cos36°
=2cos90°cos18°+cos36°
=cos36° (Proved)
[∵, cos90°=0]

Answer 2

Answer and explanation:

To prove : $4 \cos 12^\circ \cos 48^\circ \cos 72^\circ = \cos 36^\circ$

Proof :

Taking LHS,

$4 \cos 12^\circ \cos 48^\circ \cos 72^\circ$

$=(2\cos 48^\circ \cos 12^\circ)\times 2\cos 72^\circ$

Using identity, $2\cos A\cos B=\cos(A+B)+\cos(A-B)$

$=(\cos (48+12)^\circ+\cos (48-12)^\circ)\times 2\cos 72^\circ$

$=(\cos (60)^\circ+\cos (36)^\circ)\times 2\cos 72^\circ$

$=(\frac{1}{2}+\cos (36)^\circ)\times 2\cos 72^\circ$

$=\cos 72+2\cos 72^\circ\cos (36)^\circ$

$=\cos 72+\cos (72+36)^\circ+\cos (72-36)^\circ$

$=\cos 72+\cos (108)^\circ+\cos (36)^\circ$

Using identity, $\cos C+\cos D=2\cos\frac{(C+D)}{2}\cos\frac{(C-D)}{2}$

$=2\cos \frac{(108+72)}{2}\cos \frac{(108-72)}{2}+\cos (36)^\circ$

$=2\cos \frac{180}{2}\cos \frac{36}{2}+\cos (36)^\circ$

$=2\cos 90\cos 18+\cos (36)^\circ$

$=2\times 0\times \cos 18+\cos (36)^\circ$

$=\cos (36)^\circ$

=RHS

Hence proved.