Advertisements
Advertisements
Question
Prove that cos 20° cos 40° cos 60° cos 80° = `3/16`.
Advertisements
Solution
LHS = cos 20° cos 40° cos 60° cos 80°
`= cos 20° cos 40° (1/2) cos 80° [∵ cos 60° = 1/2]`
`= 1/2 (cos 20° cos 40° cos 80°)`
`= 1/2 ((2 sin 20^circ)/(2 sin 20^circ))` (cos 20° cos 40° cos 80°)
[multiply and divide by 2 sin 20°]
`= 1/2 (((2 sin 20^circ cos 20^circ)cos 40^circ cos 80^circ)/(2 sin 20^circ))`
`= 1/2 (sin (2 xx 20^circ) cos 40^circ cos 80^circ)/(2 sin 20^circ)`
`= 1/2 (sin 40^circ cos 40^circ cos 80^circ)/(2 sin 20^circ)`
`= 1/2 1/2 xx (2 sin 40^circ cos 40^circ)/(2 sin 20^circ) xx cos 80^circ`
[multiply and divide by 2]
`= 1/2 1/2 xx ((sin 2 xx 40^circ)cos 80^circ)/(2 sin 20^circ)`
`= 1/2 1/2 xx (sin 80^circ cos 80^circ)/(2 sin 20^circ)`
`= 1/2 1/2 xx 1/2 ((2 sin 80^circ cos 80^circ))/(2 sin 20^circ)`
`= 1/2 1/8 xx (sin 160^circ)/(sin 20^circ)`
`= 1/8 xx (sin (180^circ - 20^circ))/(sin 20^circ)`
`= 1/2 1/8 xx (sin 20^circ)/(sin 20^circ)`
`= 1/2 1/8 xx 1`
`= 1/2 (1/8) = 1/16`
Hence Proved.
APPEARS IN
RELATED QUESTIONS
Prove that:
sin 10° sin 50° sin 60° sin 70° = \[\frac{\sqrt{3}}{16}\]
Prove that:
\[\sin\frac{5\pi}{18} - \cos\frac{4\pi}{9} = \sqrt{3} \sin\frac{\pi}{9}\]
Prove that:
Prove that:
If cosec A + sec A = cosec B + sec B, prove that tan A tan B = \[\cot\frac{A + B}{2}\].
If cos (α + β) sin (γ + δ) = cos (α − β) sin (γ − δ), prove that cot α cot β cot γ = cot δ
If \[x \cos\theta = y \cos\left( \theta + \frac{2\pi}{3} \right) = z \cos\left( \theta + \frac{4\pi}{3} \right)\], prove that \[xy + yz + zx = 0\]
If \[m \sin\theta = n \sin\left( \theta + 2\alpha \right)\], prove that \[\tan\left( \theta + \alpha \right) \cot\alpha = \frac{m + n}{m - n}\]
Write the value of \[\frac{\sin A + \sin 3A}{\cos A + \cos 3A}\]
Prove that:
`(cos 2"A" - cos 3"A")/(sin "2A" + sin "3A") = tan "A"/2`
