原式=
sinAcosC+cosAsinC |
sinBcosC+cosBsinC |
sin(A+C) |
sin(B+C) |
sinB |
sinA |
b |
a |
∵aq+aq2>a,①
a+aq>aq2②
a+aq2>aq,③
解三个不等式可得q >
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2 |
0 <q<
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2 |
综上有
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2 |
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2 |
故答案为(
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2 |
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2 |
sinA+cosAtanC |
sinB+cosBtanC |
sinAcosC+cosAsinC |
sinBcosC+cosBsinC |
sin(A+C) |
sin(B+C) |
sinB |
sinA |
b |
a |
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2 |
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2 |
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2 |
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2 |
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2 |
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2 |