Question

30. (a) Solve: 2sin A-1 = 0. Hence, prove that sin(3A) = 3 sin A- 4sin3 A.
(b) If 2 cos A -13 = 0, find the value of A. Also, prove that sin (2A) = 2 sin A- cos A.​

Answer 1

Answer:

2Sin(A -1) = 0

SinA = 1/2

SinA = Sin30°

=> A = 30°

Now,

LHS : Sin 3A = Sin 90° = 1

RHS : 3SinA - 4Sin³A

= 3Sin30°-4Sin³30°

= 3×(1/2)-4(1/2)³

= 3/2-1/2 = 1

LHS = RHS

Hence Proved

Step-by-step explanation:

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