tanα=2,求(sin2α+cos2α)∕(1+(cosα)ˆ2)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/13 00:30:19
![tanα=2,求(sin2α+cos2α)∕(1+(cosα)ˆ2)](/uploads/image/z/10914337-1-7.jpg?t=tan%CE%B1%3D2%2C%E6%B1%82%28sin2%CE%B1%2Bcos2%CE%B1%29%E2%88%95%281%2B%28cos%CE%B1%29%26%23710%3B2%29)
xSn@*U+;3?Ʀv6A3c6AH,YTYKTUAY ?G:
vC_6s3>w^/kxpӕ']z8dâV@"#x@skPOL
T;$w(!EEJ
4ZKMFoN>?+[??]7)MЙAwۧ%{u Y_[xCqB!FeZqi/['Lѓ8 xK%zZwXd0rq0P#r%nMf "#0,F%!APJf mpİb'$2Z2ŴJ/K(
Ȳ
*gDblAwN/?άOOF[*|!HZu)b|Yg}VqV TWm:vԼ$oL[g7塔۰ܞ~:ݛv4W74`(9PP=:xzxC CreZ`R/kZ
tanα=2,求(sin2α+cos2α)∕(1+(cosα)ˆ2)
tanα=2,求(sin2α+cos2α)∕(1+(cosα)ˆ2)
tanα=2,求(sin2α+cos2α)∕(1+(cosα)ˆ2)
(sin2α+cos2α)∕(1+(cosα)ˆ2)
=(2sinacosa+cos^2a-sin^2a)/(sin^2a+2cos^2a)
=(2tana+1-tan^2a)/(tan^2a+2),(分子分母同除以cos^2a)
=(2*2+1-4)/(4+2)
=1/6
答:
tana=2,sina=2cosa
结合sin²a+cos²a=1解得cos²a=1/5
(sin2a+cos2a)/(1+cos²a)
=(2sinacosa+2cos²a-1)/(1+cos²a)
=(2*2cosa*cosa+2cos²-1)/(1+cos²a)
=(6cos²a-1)/(1+cos²a)
=6-7/(1+cos²a)
=6-7/(1+1/5)
=1/6
写起来方便,设sinα=a,cosα=b,tanα=c
原式=(2ab+a^2-b^2)/(a^2+2b^2)
母子分母同除以b^2
原式=(2c+c^2-1)/(c^2+2)=7/6