Question

If A = 30°, verify that
(i) cos 3A = 4 cos' A-3 cos A
(ii) sin 3 A = 3 sin A-4 sinº A
. Prove that
(i)
cos 45º cos 60°-sin 45°sin​

Answer 1

Step-by-step explanation:

Given, A = 30°,

(i) LHS = cos 3A = cos 3(30°) = cos 90° = 0

RHS = 4 cos^3 A - 3 cos A

= 4 cos^3 (30°) - 3 cos (30°)

= 4 (√3/2)^3 - 3 (√3/2)

= 4 ( 3√3 / 8 ) - 3√3 / 2

= 3√3 / 2 - 3√3 / 2 = 0 => LHS = RHS

.•. cos 3A = 4 cos^3 A - 3 cos A

(ii) LHS = sin 3A = sin 3(30°) = sin 90° = 1

RHS = 3 sin A - 4 sin^3 A

= 3 sin (30°) - 4 sin^3 (30°)

= 3 (1/2) - 4 (1/2)^3

= ( 3/2 ) - 4 ( 1/8 )

= 3 / 2 - 1 / 2 = 2/2 = 1

=> LHS = RHS

.•. sin 3A = 3 sin A - 4 sin^3 A

(i) cos 105° = cos 45º cos 60° - sin 45° sin 60°

we know that,

cos A cos B - sin A sin B = cos ( A + B)

hence,

cos 45º cos 60° - sin 45° sin 60° = cos ( 45° + 60° )

= cos 105°