Advertisements
Advertisements
प्रश्न
Prove that:
sin A sin(60° + A) sin(60° – A) = `1/4` sin 3A
Advertisements
उत्तर
LHS = sin A sin(60° + A) sin(60° – A)
= sin A [sin2 60° - sin2A]
[∵ sin (A + B) sin (A - B) = sin2A - sin2B]
= sin A `[(sqrt3/2)^2 - sin^2"A"]`
= sin A `[3/4 - sin^2"A"]`
= sin A `[(3 - 4 sin^2 "A")/4]`
`= 1/4` [3 sin A - 4 sin3A]
`= 1/4` sin 3A [∵ sin 3A = 3 sin A - 4 sin3A]
= RHS.
Hence proved.
APPEARS IN
संबंधित प्रश्न
Show that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
Express each of the following as the product of sines and cosines:
cos 12x + cos 8x
Prove that:
sin 50° + sin 10° = cos 20°
Prove that:
Prove that:
Write the value of \[\frac{\sin A + \sin 3A}{\cos A + \cos 3A}\]
The value of sin 50° − sin 70° + sin 10° is equal to
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
Express the following as the sum or difference of sine or cosine:
cos 7θ sin 3θ
