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sin(a-2π)sin(a+2π)-2cos(a-π|2)sin(a-π)-cos^(π|2+a) 化简
人气:194 ℃ 时间:2019-10-23 09:14:27
解答
原式=sin[-(2π-a)]sin(2π+a) - 2cos[-(π/2 - a)]sin[-(π-a)] - cos²(π/2 + a)
=[-sin(2π-a)]sina - 2cos(π/2 - a)[-sin(π-a)] - (-sina)²
=[-(-sina)]sina - 2sina(-sina) - sin²a
=sin²a + 2sin²a - sin²a
=2sin²a
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