1.化简(1-tan^4B)cos^2B+tan^2B(要过程) 2.tanA=-1\2,则2sinAcosA\sin^
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1.化简(1-tan^4B)cos^2B+tan^2B(要过程) 2.tanA=-1\2,则2sinAcosA\sin^2A-cos^2A=?
3.若cosA=2sinA,则1+cos^2A\sinAcosA
4.若 sin^2A+sinA=1,则cos^2A+cos^4A=?
3.若cosA=2sinA,则1+cos^2A\sinAcosA
4.若 sin^2A+sinA=1,则cos^2A+cos^4A=?
楼主的问题比较多,呵呵,一个个的说吧.
1,化简:(1-sin^4B/cos^4B)cos^2B+sin^2B/cos^2B=(cos^4B-sin^4B)/cos^2B+sin^2B/cos^2B=((cos^2B+sin^2B)(cos^2B-sin^2B)+sin^2B)/cos^2B=(cos^2B-sin^2B+sin^2B)/cos^2B=1;
2,你只要对分子和分母同除cos^2A,再结合已知,即可求出你想要的答案.
3,也是一样分子和分母同除cos^2A,由已知条件可知道tanA,和cotA的值来,也是很快得到结果.
4,由已知可得出cos^2A=sinA,容易的出结果是1;
1,化简:(1-sin^4B/cos^4B)cos^2B+sin^2B/cos^2B=(cos^4B-sin^4B)/cos^2B+sin^2B/cos^2B=((cos^2B+sin^2B)(cos^2B-sin^2B)+sin^2B)/cos^2B=(cos^2B-sin^2B+sin^2B)/cos^2B=1;
2,你只要对分子和分母同除cos^2A,再结合已知,即可求出你想要的答案.
3,也是一样分子和分母同除cos^2A,由已知条件可知道tanA,和cotA的值来,也是很快得到结果.
4,由已知可得出cos^2A=sinA,容易的出结果是1;
sinAcosA/(sin^2A+cos^2A) =tanA/(1+tan^2A),为什么
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