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
Prove that:
cos 40° cos 80° cos 160° = \[- \frac{1}{8}\]
Prove that tan 20° tan 30° tan 40° tan 80° = 1.
Show that:
sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0
Prove that:
sin 40° + sin 20° = cos 10°
Prove that:
Prove that:
If sin 2A = λ sin 2B, then write the value of \[\frac{\lambda + 1}{\lambda - 1}\]
If sin α + sin β = a and cos α − cos β = b, then tan \[\frac{\alpha - \beta}{2}\]=
sin 47° + sin 61° − sin 11° − sin 25° is equal to
