①(2cosα+3sinα)/(3cosα+sinα) (2+3tanα)/(3+tanα)2(sinα)^2+sinαcosα-3(cosα)^2②tanα√((1/((sinα)^2))-1)第二个可以看图:
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![①(2cosα+3sinα)/(3cosα+sinα) (2+3tanα)/(3+tanα)2(sinα)^2+sinαcosα-3(cosα)^2②tanα√((1/((sinα)^2))-1)第二个可以看图:](/uploads/image/z/4336205-5-5.jpg?t=%E2%91%A0%282cos%CE%B1%2B3sin%CE%B1%29%2F%283cos%CE%B1%2Bsin%CE%B1%29+%282%2B3tan%CE%B1%29%2F%283%2Btan%CE%B1%292%28sin%CE%B1%29%5E2%2Bsin%CE%B1cos%CE%B1-3%28cos%CE%B1%29%5E2%E2%91%A1tan%CE%B1%E2%88%9A%28%281%2F%28%28sin%CE%B1%29%5E2%29%29-1%29%E7%AC%AC%E4%BA%8C%E4%B8%AA%E5%8F%AF%E4%BB%A5%E7%9C%8B%E5%9B%BE%EF%BC%9A)
①(2cosα+3sinα)/(3cosα+sinα) (2+3tanα)/(3+tanα)2(sinα)^2+sinαcosα-3(cosα)^2②tanα√((1/((sinα)^2))-1)第二个可以看图:
①(2cosα+3sinα)/(3cosα+sinα) (2+3tanα)/(3+tanα)
2(sinα)^2+sinαcosα-3(cosα)^2
②tanα√((1/((sinα)^2))-1)
第二个可以看图:
①(2cosα+3sinα)/(3cosα+sinα) (2+3tanα)/(3+tanα)2(sinα)^2+sinαcosα-3(cosα)^2②tanα√((1/((sinα)^2))-1)第二个可以看图:
(2cosα+3sinα)/(3cosα+sinα) 分子分母同时除以cosa就可得到答案
2(sinα)^2+sinαcosα-3(cosα)^2
因为
sinαcosα=1/2 sin2a
2(sinα)^2= - (1-2(sina)^2)+1= - cos2a+1
-3(cosα)^2=-3/2(2(cosα)^2-1)-3/2=-3/2 cos2a-3/2
所以原式=1/2 sin 2a-5/2 cos 2a-1/2=√26/2sin(2a-b) (tanb=5)
tanα√((1/((sinα)^2))-1)
根号内:通分得
(1-(sina)^2)/(sina)^2,因为1=(sina)^2+(cosa)^2
所以得(cosa)^2/(sina)^2
带上根号得[cota]
所以原式=1或-1
(1)(2cosα+3sinα)/(3cosα+sinα)
=(2cosα/cosα+3sinα/cosα)/(3cosα/cosα+sinα/cosα)
=(2+3tanα)/(3+tanα)
2(sinα)^2+sinαcosα-3(cosα)^2
=(2sinα+3cosα)(sinα-cosα)
(2)tanα√((1/((sinα)^2))-1)
=tanα√[(1-(sinα)^2)/(sinα)^2]
=tanα√[(cosα)^2/(sinα)^2]
=tanα*|cotα|
=1或-1
①(2cosα+3sinα)/(3cosα+sinα) 分子分母同除以cosα,结果就是答案了。 2(sinα)^2+sinαcosα-3(cosα)^2 =1/2sin2α+2-5(cosα)^2 ∵2(cosα)^2-1=cos2α ∴1/2sin2α+2-5(cosα)^2=1/2sin2α-5/2cos2α-1/2=1/2(sin2α-5cos2α-1) ②tanα√((1/((sinα)^2))-1)
第一个这样:
首先(2cosα+3sinα)/(3cosα+sinα) 分子分母同时除以cosa就可得到答案
2(sinα)^2+sinαcosα-3(cosα)^2
因为
sinαcosα=1/2 sin2a
2(sinα)^2= - (1-2(sina)^2)+1= - cos2a+1
-3(cosα)^2=-3/2(2(cosα)^2-1)-3...
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第一个这样:
首先(2cosα+3sinα)/(3cosα+sinα) 分子分母同时除以cosa就可得到答案
2(sinα)^2+sinαcosα-3(cosα)^2
因为
sinαcosα=1/2 sin2a
2(sinα)^2= - (1-2(sina)^2)+1= - cos2a+1
-3(cosα)^2=-3/2(2(cosα)^2-1)-3/2=-3/2 cos2a-3/2
所以原式=1/2 sin 2a-5/2 cos 2a-1/2=√26/2sin(2a-b) (tanb=5)
tanα√((1/((sinα)^2))-1)
根号内:通分得
(1-(sina)^2)/(sina)^2,因为1=(sina)^2+(cosa)^2
所以得(cosa)^2/(sina)^2
带上根号得[cota]
所以原式=1或-1
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