cos(a-3π)/sin(a+5π)*sin(a-2π)*cos(2π-a)
=(-cosa)/(-sina)*sinacosa
=cos²a已知cos(π+a)=4/5,求cos(π+a)/cosa【cos(π-a)-1】+cos(a-2π)/cosacos(π-a)+cos(a-2π)采納我,重新問
化簡cos(a-3π)/sin(a+5π)*sin(a-2π)
化簡cos(a-3π)/sin(a+5π)*sin(a-2π)
化簡cos(a-3π)/sin(a+5π)*sin(a-2π)*cos(2π-a)
化簡cos(a-3π)/sin(a+5π)*sin(a-2π)*cos(2π-a)
數(shù)學(xué)人氣:960 ℃時間:2020-05-14 08:21:06
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