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sin(a-3pai)cos(pai-a) / cos(pai/2 - a)cos(pai+a)sin(pai/2 + a)
人气:364 ℃ 时间:2020-04-24 02:34:16
解答
sin(a-3π)cos(π-a) / cos(π/2 - a)cos(π+a)sin(π/2 + a)
=-sina[-cos(-a)] / sina[-cosa][-sin(π/2 + a-π)]
=sinacosa / sina cosa sin(-π/2+a)
=sinacosa / sina cosa [-sin(π/2-a)]
=sinacosa / sina cosa (-cosa)
=-1/cosa
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