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| Theorem 1 |
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| General solution of sin q = k |
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| Method I |
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| Given sin q = k .......(i) |
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| Let a be any angle, such that sin a = k. |
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| (In actual practice, we choose the smallest positive angle such that sin a = k.) |
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| From (i) and (ii), we have |
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| Method II |
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| Given equation sin q = k. |
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| Let a be the least positive angle, such that sin a = k. |
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| Also, any angle co-terminal with the angles a, p - a is trigonometrically equivalent and so has the same 'sine'. |
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| Particular Cases |
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| i) If sin q = 0, then a = 0. |
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| Hence, we get q = np + (-1)n 0 = np. |
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