Question

Cos*3(2a) +3cos2a=4 (cos*6a- sin*6a)

Answer 1

Given: Cos*3(2a) +3cos2a=4 (cos*6a- sin*6a)

To find: The value of the above expression?

Solution:

• Consider RHS, we have 4 (cos*6a- sin*6a)

• Now we have the formula for power reduction, which is:

              cos^6 a = (1+cos2a / 2 )³

                            = 1/8 x ( 1 + 3 cos2a + 3 cos²2a + cos³2a)

• Similarly,

              sin^6 a = (1-cos2a / 2 )³

                           = 1/8 x ( 1 - 3 cos2a + 3 cos²2a - cos³2a)

• Adding both the terms, we get:

              cos*6a- sin*6a = 1/8 ( 1 + 3 cos2a + 3 cos²2a + cos³2a + 1 - 3 cos2a + 3 cos²2a - cos³2a )

• Mul;tiplying by 4, we get RHS:

              4 (cos*6a- sin*6a) = 1/2 ( 6 cos2a + 2 cos³2a)

              4 (cos*6a- sin*6a) = 3 cos2a + cos³2a

                                            = LHS

Answer:

                   So from above solution we proved that Cos*3(2a) +3cos2a=4 (cos*6a- sin*6a)