Advertisements
Advertisements
प्रश्न
Prove that:
cos 20° cos 40° cos 80° = `1/8`
Advertisements
उत्तर
cos 20° cos 40° cos 80°
= `((2 sin 20^circ)/(2 sin 20^circ))` cos 20° cos 40° cos 80°
[multiply and divide by 2 sin 20°]
`= ((2 sin 20^circ cos 20^circ) cos 40^circ cos 80^circ)/(2 sin 20^circ)`
`= (sin (2 xx 20^circ) cos 40^circ cos 80^circ)/(2 sin 20^circ)`
= `(sin 40^circ cos 40^circ cos 80^circ)/(2 sin 20^circ)`
(Multiply and divide by 2)
`= 1/2 xx ((2 sin 40^circ cos 40^circ))/(2 sin 20^circ) cos 80^circ`
`= 1/2 xx ((sin 2 xx 40^circ)cos 80^circ)/(2 sin 20^circ)`
`= 1/2 xx (sin 80^circ cos 80^circ)/(2 sin 20^circ)`
`= 1/2 xx 1/2 ((2 sin 80^circ cos 80^circ))/(2 sin 20^circ)`
`= 1/8 xx (sin 160^circ)/(sin 20^circ)`
`= 1/8 xx sin (180^circ - 20^circ)/(sin 20^circ)`
`= 1/8 xx sin 20^circ/sin 20^circ` ...[∵ sin(180° – θ) = sin θ]
`= 1/8 xx 1 = 1/8`
APPEARS IN
संबंधित प्रश्न
Prove that:
sin 38° + sin 22° = sin 82°
Prove that:
cos 80° + cos 40° − cos 20° = 0
Prove that:
If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), then write the value of tan A tan B tan C.
Express the following as the sum or difference of sine or cosine:
`sin "A"/8 sin (3"A")/8`
Express the following as the product of sine and cosine.
sin A + sin 2A
Prove that:
(cos α – cos β)2 + (sin α – sin β)2 = 4 sin2 `((alpha - beta)/2)`
Prove that:
2 cos `pi/13` cos \[\frac{9\pi}{13} + \text{cos} \frac{3\pi}{13} + \text{cos} \frac{5\pi}{13}\] = 0
If tan θ = `1/sqrt5` and θ lies in the first quadrant then cos θ is:
If secx cos5x + 1 = 0, where 0 < x ≤ `pi/2`, then find the value of x.
