# What is a if sin 3a*cos 5a=sin 6a*cos 2a?

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You need to convert the products into sums using the following formula such that:

`sin alpha*cos beta = (1/2)(sin(alpha - beta) + sin(alpha + beta))`

Reasoning by analogy yields:

`sin 3a*cos 5a = (1/2)(sin(3a - 5a) + sin(3a + 5a))`

`sin 3a*cos 5a = (1/2)(sin(-2a) + sin(8a))`

You need to remember that the sine function is odd, hence `sin(-2a) = -sin 2a` such that:

`sin 3a*cos 5a = (1/2)(sin 8a - sin 2a)`

`sin 6a*cos 2a = (1/2)(sin(6a - 2a) + sin(6a + 2a))`

`sin 6a*cos 2a = (1/2)(sin 8a + sin 4a)`

You need to substitute `(1/2)(sin 8a + sin 4a)` for `sin 6a*cos 2a` and `(1/2)(sin 8a - sin 2a) for sin 3a*cos 5a` such that:

`(1/2)(sin 8a - sin 2a) = (1/2)(sin 8a + sin 4a)`

Reducing by `1/2` yields:

`sin 8a - sin 2a = sin 8a + sin 4a`

Reducing by `sin 8a ` yields:

`- sin 2a = sin 4a => sin (-2a) = sin 4a`

`-2a = (-1)^n*4a + npi`

If `n = 2k => -2a = 4a + 2kpi => -6a = 2kpi => a = -2kpi/6`

If `n = 2k+1 => -2a = -4a + 2kpi => 2a = 2kpi => a =kpi`

**Hence, evaluating the general solutions to the given equation yields `a =kpi` and `a = -2kpi/6` .**