Advertisements
Advertisements
Question
Prove that cos 20° cos 40° cos 60° cos 80° = `3/16`.
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:
Prove that:
Prove that:
sin 50° + sin 10° = cos 20°
Prove that:
cos 80° + cos 40° − cos 20° = 0
Prove that:
If \[x \cos\theta = y \cos\left( \theta + \frac{2\pi}{3} \right) = z \cos\left( \theta + \frac{4\pi}{3} \right)\], prove that \[xy + yz + zx = 0\]
sin 163° cos 347° + sin 73° sin 167° =
If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P., then cot A, cot B and cot Care in
Prove that:
`(cos 7"A" +cos 5"A")/(sin 7"A" −sin 5"A")` = cot A
If sin(y + z – x), sin(z + x – y), sin(x + y – z) are in A.P, then prove that tan x, tan y and tan z are in A.P.
