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A.math - trigonometry identities
Prove the following identities. (a) (sin a)^4 - (cos a)^4 = 2 (sin a)^2 - 1 (b) (1+cos a+sin a) / (1+cos a - sin a) = (1+sin a) / cos a (c) tan a?(1 - sin a) / 1+cos a) = cot a?(1 - cos a) / (1+sin a)
最佳解答:
(a) LHS = sin?a - cos?a = (sin2a - cos2a) (sin2a + cos2a) = [sin2a - (1 - sin2a)] (1) = 2sin2a - 1 = RHS (b) LHS = (1+cos a+sin a) / (1+cos a - sin a) = (1+cos a+sin a) / (1+cos a - sin a) x (1+cos a+sin a) / (1+cos a+sin a) = [1 + 2cos a + cos2a + 2sin a(1+cos a) + sin2a] / (1 + 2cos a + cos2a - sin2a) = [2 + 2cos a + 2sin a(1+cos a)] / (2cos a + 2cos2a) = [2(1 + cos a) + 2sin a(1+cos a)] / [2 cos a(1 + cosa)] = (1 + sin a) / cos a = RHS (c) LHS = tan a?(1 - sin a) / (1 + cos a) = [tan a?(1 - sin a) / (1 + cos a)]?[(1 + sin a)(1 - cos a)] / [(1 + sin a)(1 - cos a)] = tan a?(1 - sin2a)(1 - cos a) / [(1 - cos2a) (1 + sin a)] = tan a?cos2a(1 - cos a) / [sin2a(1 + sin a)] = cot a?(1 - cos a) / (1 + sin a) = RHS
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A.math - trigonometry identities
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發問:Prove the following identities. (a) (sin a)^4 - (cos a)^4 = 2 (sin a)^2 - 1 (b) (1+cos a+sin a) / (1+cos a - sin a) = (1+sin a) / cos a (c) tan a?(1 - sin a) / 1+cos a) = cot a?(1 - cos a) / (1+sin a)
最佳解答:
(a) LHS = sin?a - cos?a = (sin2a - cos2a) (sin2a + cos2a) = [sin2a - (1 - sin2a)] (1) = 2sin2a - 1 = RHS (b) LHS = (1+cos a+sin a) / (1+cos a - sin a) = (1+cos a+sin a) / (1+cos a - sin a) x (1+cos a+sin a) / (1+cos a+sin a) = [1 + 2cos a + cos2a + 2sin a(1+cos a) + sin2a] / (1 + 2cos a + cos2a - sin2a) = [2 + 2cos a + 2sin a(1+cos a)] / (2cos a + 2cos2a) = [2(1 + cos a) + 2sin a(1+cos a)] / [2 cos a(1 + cosa)] = (1 + sin a) / cos a = RHS (c) LHS = tan a?(1 - sin a) / (1 + cos a) = [tan a?(1 - sin a) / (1 + cos a)]?[(1 + sin a)(1 - cos a)] / [(1 + sin a)(1 - cos a)] = tan a?(1 - sin2a)(1 - cos a) / [(1 - cos2a) (1 + sin a)] = tan a?cos2a(1 - cos a) / [sin2a(1 + sin a)] = cot a?(1 - cos a) / (1 + sin a) = RHS
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