# prove sin2a+sin2b-sin2c=4cosacosbcosc

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You need to use a condition between the angles a,b,c to prove the given identity, hence, supposing that `a+b = pi - c` yields:

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

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

Substituting sin c for `sin(a+b)` yields:

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

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

You need to use the double angle identity for `sin 2c` such that:

`sin 2c = 2 sin c cos c`

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

Factoring out `2sin c` yields:

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

You need to convert the difference `cos (a - b) - cos c` into a product, such that:

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

Substituting `pi - b` for `a + c ` and `pi - a` for `c + b` yields:

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

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

Using the identity `sin (pi/2 - alpha) = cos alpha` yields:

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

**Hence, checking if the given expression is identity yields that `sin 2a + sin 2b - sin 2c = 4 sin c* cos b* cos a != sin 2a + sin 2b - sin 2c = 4cos c* cos b* cos a` , thus, the expression is not an identity.**