5sinB=sin(2A+B) 求证2tan(A+B)=3tanA
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5sinB=sin(2A+B) 求证2tan(A+B)=3tanA
![5sinB=sin(2A+B) 求证2tan(A+B)=3tanA](/uploads/image/z/19225146-66-6.jpg?t=5sinB%3Dsin%282A%2BB%29+%E6%B1%82%E8%AF%812tan%28A%2BB%29%3D3tanA)
原式可以写成:
5sin[(A + B) - A] = sin[(A + B) + A]
左边 = 5sin(A + B)cosA - 5cos(A + B)sinA
右边 = sin(A + B)cosA + cos(A + B)sinA ,移项得:
4sin(A + B)cosA = 6cos(A + B)sinA
两边除以 2cosAcos(A + B) 即得:
2tan(A+B)=3tanA
5sin[(A + B) - A] = sin[(A + B) + A]
左边 = 5sin(A + B)cosA - 5cos(A + B)sinA
右边 = sin(A + B)cosA + cos(A + B)sinA ,移项得:
4sin(A + B)cosA = 6cos(A + B)sinA
两边除以 2cosAcos(A + B) 即得:
2tan(A+B)=3tanA
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