Advertisements
Advertisements
Question
Prove that:
sin A sin(60° + A) sin(60° – A) = `1/4` sin 3A
Advertisements
Solution
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
RELATED QUESTIONS
Show that:
sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0
Prove that:
cos 20° + cos 100° + cos 140° = 0
Prove that:
cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
Prove that:
`sin A + sin 2A + sin 4A + sin 5A = 4 cos (A/2) cos((3A)/2)sin3A`
Prove that:
cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = −\[\frac{3}{4}\]
Prove that:
Prove that:
Write the value of \[\sin\frac{\pi}{15}\sin\frac{4\pi}{15}\sin\frac{3\pi}{10}\]
Prove that:
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
