Question

cos36cos72cos108cos144 =1/16

Answer 1

LHS = cos36°.cos72°.cos108°.cos144°

= cos36°.cos72°.cos(180°-72°).cos(180°-36°)

= cos36°.cos72°.(-cos72°).(-cos36°)

= (cos36°.cos72°)²

we know,
$cos36=\frac{1}{4}(\sqrt{5}+1)$
$cos72=\frac{1}{4}(\sqrt{5}-1)$

= $\{\frac{1}{4}(\sqrt{5}+1).\frac{1}{4}(\sqrt{5}-1)\}^2$

= {1/16(5 - 1)}²

= {1/4}² = 1/16 = RHS

Answer 2

HELLO DEAR,

GIVEN:-

cos36 * cos72 * cos108 * cos144 = 1/16

solving L.H.S,

cos36 * cos72 * cos108 * cos144 = k(say)

multiply both side by "2sin36",

2ksin36 = (2sin36 * cos36) * cos72 * cos108 * cos144

2ksin36 = sin72 * cos 72 * cos108 * cos144

2ksin36 = 1/2{2(sin72 * cos72) * cos108 * cos144}

2ksin36 = 1/4{(2sin144 * cos144) * cos108}

2ksin36 = 1/4{sin288 * cos108}

2ksin36 = 1/8{2sin(180 + 108) * cos108}

2ksin36 = 1/8{-2sin108 * cos 108}

[as , sin(180 + $\Theta$) = -sin$\Theta$]

2ksin36 = 1/8{-sin216}

2ksin36 = -1/8{sin(180 + 36)}

2ksin36 = -1/8{-sin36}

2ksin36 = 1/8(sin36)

k = 1/16

HENCE, cos36cos72cos108cos144 = 1/16

I HOPE ITS HELP YOU DEAR,

THANKS