Question

cos⁻¹(4x³-3x) = 3cos⁻¹x, 1/2 < x < 1,prove it

Answer 1

Let us consider 3cos^-1x = A .......(1)

cos^-1x = A/3 => cosA/3= x ..........(2)

we know, $\bf{cos3\theta=4cos^3\theta-3cos\theta}$

so, cosA = 4cos³A/3 - 3cosA/3

= 4x³ - 3x [ from equation (2), ]

cosA = 4x³ - 3x

A = cos^-1 (4x³ - 3x)

from equation (1),

cos^(4x³ - 3x) = 3cos^-1x
hence proved//

Answer 2

HELLO DEAR,



GIVEN:-
cos-¹(4x³-3x) = 3cos-¹x,

L.H.S =>
cos-¹(4x³-3x)
put x = cost

so, cos-¹ (4cos³t - 3t) =

we know:- cos3∅ = 4cos³∅ - 3cos∅

therefore, cos-¹ (cos3t) = 3t

[as , cos-¹(cos p) = p ]

=> 3t

[as , x = cost => t = cos-¹]

=> 3cos-¹x = R.H.S


HENCE, L.H.S = R.H.S


I HOPE ITS HELP YOU DEAR,
THANKS