# Prove if cosine(a+b+c)=1, then sin(2a+b+c)=sina?

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You need to prove that `sin(2a+b+c) = sin a,` hence, moving sin a to the left side, you may convert the difference `sin(2a+b+c) - sin a ` into a product, such that:

`sin(2a+b+c) - sin a = 0 => 2 cos((2a + b + c + a)/2)*sin ((2a + b + c - a)/2) = 0`

`2 cos((3a + b + c)/2)*sin ((a + b + c)/2) = 0`

The problem provides the information that `cos(a + b + c) = 1` , hence, `a + b + c = 0` or `a + b + c = 2pi.`

Replacing 0 for `a + b + c` yields:

`2 cos((2a)/2)*sin ((0)/2) = 0`

`2cos a*sin 0 = 0`

Since `sin 0 = 0` , hence, using zero product rule yields ` 2cos a*sin 0 = 0` is valid.

Replacing `2pi ` for` a + b + c` yields:

`2 cos((2a + 2pi)/2)*sin ((2pi)/2) = 0`

`2 cos(a + pi)*sin (pi) = 0`

Since `sin pi = 0` , hence, using zero product rule yields `2 cos(a + pi)*sin (pi) = 0` is valid.

**Hence, testing if `sin(2a+b+c) = sin a` , under the given conditions, yields the valid statements `2cos a*sin 0 = 0` or **`2 cos(a + pi)*sin (pi) = 0.`

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