赤峰二中11级8班吧 关注：33贴子：1,147
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Very interesting question!
Okay, multiplying them together gives
ab = sin(x)cos(x) + sin(y)cos(y) + sin(x)cos(y) + sin(y)cos(x)
=½sin(2x) + ½sin(2y) + sin(x + y).
Using the sum to product formula sin(A) + sin(B) = 2sin(½(A+B))cos(½(A-B)) on the first two terms gives
ab = sin(x + y)cos(x - y) + sin(x + y) = sin(x + y)(cos(x - y) + 1).
If we square both equations, we get
sin²(x) + sin²(y) + 2sin(x)sin(y) = a²
cos²(x) + cos²(y) + 2cos(x)cos(y) = b².
Adding these lines together gives
a² + b² = 2 + 2cos(x)cos(y) + 2cos(x)cos(y) = 2 + 2cos(x - y).
So cos(x - y) = ½(a² + b²) - 1. Substituting this back into the first equation gives
ab = sin(x + y)(cos(x - y) + 1) = sin(x + y)(½(a² + b²) - 1 + 1) ==>
sin(x + y)(½(a² + b²)) = ab.
If at least one of a or b is zero, then sin(x + y) = 0. Otherwise
sin(x + y) = 2ab/(a² + b²).
Whew!
Okay, multiplying them together gives
ab = sin(x)cos(x) + sin(y)cos(y) + sin(x)cos(y) + sin(y)cos(x)
=½sin(2x) + ½sin(2y) + sin(x + y).
Using the sum to product formula sin(A) + sin(B) = 2sin(½(A+B))cos(½(A-B)) on the first two terms gives
ab = sin(x + y)cos(x - y) + sin(x + y) = sin(x + y)(cos(x - y) + 1).
If we square both equations, we get
sin²(x) + sin²(y) + 2sin(x)sin(y) = a²
cos²(x) + cos²(y) + 2cos(x)cos(y) = b².
Adding these lines together gives
a² + b² = 2 + 2cos(x)cos(y) + 2cos(x)cos(y) = 2 + 2cos(x - y).
So cos(x - y) = ½(a² + b²) - 1. Substituting this back into the first equation gives
ab = sin(x + y)(cos(x - y) + 1) = sin(x + y)(½(a² + b²) - 1 + 1) ==>
sin(x + y)(½(a² + b²)) = ab.
If at least one of a or b is zero, then sin(x + y) = 0. Otherwise
sin(x + y) = 2ab/(a² + b²).
Whew!
