# sin 2a + sin 2b - sin 2c = 4 cos a cos b . sin c

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### 1 Answer

You need to consider that a,b and c are the internal angles of a triangle, hence, the sum of internal angles is of `180^o` (pi radians) such that:

`a + b + c = pi =gt a + b = pi - c`

You need to use sine function to evaluate `a + b` and `pi - c` such that:

`sin(a+b) = sin(pi-c)`

`sin(a+b) = sin pi*cos c - sin c*cos pi`

Since `sin pi= 0` and `cos pi = -1` , yields:

`sin(a+b) = sin c`

You need to convert the sum `sin 2a + sin 2b ` into a product such that:

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

`sin 2a + sin 2b = 2 sin(a+b)*cos(a-b)`

You need to substitute `2 sin(a+b)*cos(a-b)` for `sin 2a + sin 2b` in identity to be proved such that:

`2 sin(a+b)*cos(a-b) - sin 2c = 4 cos a*cos b*sin c`

Notice that `sin(a+b) = sinc` , hence, you should substitute `sin c` for `sin(a+b)` such that:

2 sin c*cos(a-b) - sin 2c = 4 cos a*cos b*sin c You need to use the double angle formula for `sin 2c ` such that:

`2 sin c*cos(a-b) -2 sin c* cos c= 4 cos a*cos b*sin c`

You need to factor out `2 sin c` to the left side such that:

`2 sin c*(cos(a-b)-cos c) = 4 cos a*cos b*sin c`

You need to divide by `2 sin c` both sides such that:

`(cos(a-b)-cos c) =2 cos a*cos b`

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

You need to use again the identity `a+b+c = pi` such that:

`a + c = pi - b`

`sin ((a - b + c)/2) = sin (pi/2 - 2b/2) = sin (pi/2 - b)`

You need to remember that `sin (pi/2 - b) = cos b` siuch that:

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

`b + c = pi - a =gt sin ((c - a + b)/2) = sin ((pi - 2a)/2) = cos a`

Substituting `cos a` for `sin ((c - a + b)/2)` yields:

`2 cos b*cos a= 2 cos a*cos b`

**Notice that the last line proves that the expression `sin 2a + sin 2b - sin 2c = 4 cos a*cos b*sin c` is an identity for `a+b+c = pi.` **

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