若函数f(x)的解析式为f(x)=2tanx-(2sin^2 (x/2) -1)/(sinx/2*c

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1）f(x)=2tanx-(2sin^2 (x/2) -1)/(sinx/2*cosx/2)
f(x)=2tanx +cos x /(1/2 * sin x )
=2tanx +2 cotx
=2[(sin x /cos x )+(cos x /sin x)]=4/sin(2x)
将π/12带入,=8
1）求sin^2+sin2α/cos^2 α +cos2α
sinα=4/5 ,cosα=3/5,
(sinα)^2=16/25,(cosα)^2=9/25
cos2α=2(cosα)^2-1=-7/25
sinα=2sinαcosα=24/25
(sin^2+sin2α)/(cos^2 α +cos2α)
=(16/25+24/25)/(9/25-7/25)=40/2=20
2）tan(α-5π/4)
tan(α-5π/4)
=(tanα -tan5π/4)/(1+tanα*tan5π/4),
tan5π/4=1,sinα=4/5 ,cosα=3/5,tanα=4/3,
带入tan(α-5π/4)中
=(tanα -tan5π/4)/(1+tanα*tan5π/4)
=(4/3-1)/(1+4/3)
=1/7======以下答案可供参考======
供参考答案1:
1）f(x)=2tanx-(2sin^2 (x/2) -1)/(sinx/2*cosx/2)
f(x)=2tanx +cos x /(1/2 * sin x )
=2tanx +2 cotx
=2[(sin x /cos x )+(cos x /sin x)]=4/sin(2x)
将π/12带入,=8
2）已知α为锐角,
1）求sin^2+sin2α/cos^2 α +cos2α
sinα=4/5 ,cosα=3/5,
(sinα)^2=16/25,(cosα)^2=9/25
cos2α=2(cosα)^2-1=-7/25
sinα=2sinαcosα=24/25
(sin^2+sin2α)/(cos^2 α +cos2α)
=(16/25+24/25)/(9/25-7/25)=40/2=20
2）tan(α-5π/4)
tan(α-5π/4)
=(tanα -tan5π/4)/(1+tanα*tan5π/4),
tan5π/4=1,sinα=4/5 ,cosα=3/5,tanα=4/3,
带入tan(α-5π/4)中
=(tanα -tan5π/4)/(1+tanα*tan5π/4)=(4/3-1)/(1+4/3)=1/7供参考答案2:1。若函数f(x)的解析式为f(x)=2tanx-(2sin^2 (x/2) -1)/(sinx/2*cosx/2) 则f(π/12)是f(x)=2tg x +cos x /(1/2 * sin x )=2tg x +2 ctg x =2[(sin x /cos x )+(cos x /sin x)]=2/[(1/2)sin 2x]将π/12带入,可算的答案为82。已知α为锐角,且sinα=4/5 （1）求sin^2+sin2α/cos^2 α +cos2α的值 题目不完整,没法给你算（2）tan(α-5π/4) ,tan(α-5π/4)=(tanα -tan5π/4)/(1+tanα*tan5π/4),因为tan5π/4=tanπ/4=1,又因为sinα=4/5 ,所以cosα=3/5,所以tanα=4/3,带入tan(α-5π/4)=(tanα -tan5π/4)/(1+tanα*tan5π/4)=(4/3-1)/(1+4/3)=1/7