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已知tana=2,则2sin^a+sinacosa
人气:351 ℃ 时间:2020-05-23 16:31:39
解答
sin2a=2tana/[1+(tana)^2]
=2*2/(1+2^2)
=4/5
cos2a=[1-(tana)]^2/[1+(tana)^2]
=(1-2^2)/(1+2^2)
=-3/5
2(sina)^2+sinacosa
=2*[1-cos2a]/2+1/2*sin2a
=1-cos2a+1/2*sin2a
=1/2*sin2a-cos2a+1
=1/2*4/5-(-3/5)+1
=2/5+3/5+1
=1+1
=2
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