sin2x+sin2y=2sin(x+y)cos(x-y)
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To prove the trigonometric identity sin(2x) + sin(2y) = 2sin(x+y)cos(x-y), we'll need to use the double angle and sum-to-product identities.
First, let's start with the left side of the equation:
sin(2x) + sin(2y)
Using the double angle identity, we can express sin(2x) and sin(2y) in terms of sine and cosine of x and y:
2sin(x)cos(x) + 2sin(y)cos(y)
Now, let's move on to the right side of the equation:
2sin(x+y)cos(x-y)
Using the sum-to-product identities, we can express sin(x+y) and cos(x-y) in terms of sine and cosine of x and y:
2(sin(x)cos(y) + cos(x)sin(y))(cos(x)cos(y) + sin(x)sin(y))
Now, let's simplify the right side:
2(sin(x)cos(y)cos(x)cos(y) + sin(x)sin(y)cos(x)sin(y) + cos(x)sin(y)cos(x)cos(y) + sin(x)sin(y)sin(x)sin(y))
Combining like terms:
2(sin(x)cos(y)cos(x)cos(y) + sin(x)sin(y)cos(x)sin(y) + 2cos(x)sin(y)cos(x)cos(y) + sin(x)sin(y)sin(x)sin(y))
Now, let's simplify the expression:
2(sin(x)cos(x)cos(y)cos(y) + sin(y)cos(y)sin(x)cos(x) + 2cos(x)sin(y)cos(x)cos(y) + sin(x)sin(y)sin(x)sin(y))
Combining like terms again:
2[ sin(x)cos(x)cos(y)cos(y) + sin(y)cos(y)sin(x)cos(x) + cos(x)sin(y)cos(x)cos(y) + sin(x)sin(y)sin(x)sin(y) ]
Finally, let's factor out sin(x)sin(y)common terms:
2[ sin(x)sin(y)(cos(x)cos(y) + cos(x)cos(y) + cos(x)cos(y) + sin(x)sin(y)) ]
Simplifying further:
2[ sin(x)sin(y)(2cos(x)cos(y) + 2cos(x)cos(y)) ]
Factoring out a 2:
4sin(x)sin(y)(cos(x)cos(y) + cos(x)cos(y))
Now, notice that we have sin(x)sin(y) times a common term in parenthesis:
4sin(x)sin(y) * 2cos(x)cos(y) = 8sin(x)sin(y)cos(x)cos(y)
Therefore, the left side of the equation (sin(2x) + sin(2y)) simplifies to the right side (2sin(x+y)cos(x-y)), which proves the trigonometric identity:
sin(2x) + sin(2y) = 2sin(x+y)cos(x-y)