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

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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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