Question

if 4 cos²x -3 =0 and 0<=X<=90 prove that sin 3x = 3 sin x - 4 sin³x​

Answer 1

Answer:

Step-by-step explanation:

  4 cos²x  - 3 =0

  cos²x  = $\frac{3}{4}$

  cosx = $\frac{\sqrt{3} }{2}$

        x = 30° ( from trigonometric angle table)

now,

L.H.S. = sin3x = sin3(30) = sin90 = 1

R.H.S. = 3 sin x - 4 sin³x​

            = 3sin(30) - 4 sin³ (30)

            = (3 x $\frac{1}{2}$ ) - 4($(\frac{1}{2})^{3}$

            = $\frac{3}{2} - 4*\frac{1}{8}$

            = $\frac{3}{2} - \frac{1}{2}$

            = 1

LHS = RHS

Hence sin 3x = 3 sin x - 4 sin³x​  proved....