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三角函数 (19 15:4:51)
已知tanx=1/3,求(1)(5sinx+3cosx)/(6cosx-4sinx);(2)sin2x-3sinxcosx+2cos2x的值.
人气:125 ℃ 时间:2020-05-07 07:59:53
解答
(1)(5sinx+3cosx)/(6cosx-4sinx);
=(5tanx+3)/(6-4tanx),【分子分母同除以cosx】
=(5/3+3)/(6-4/3)
=1
tan2x=2tanx/(1-tan^x)=(2/3)/(1-1/9)=3/4
sin2x=3/5
cos2x=4/5
(2)原式=2sinxcosx-3sinxcosx+2cos2x
=-sinxcosx+2cos2x
=-1/2sin2x+2cos2x
=-1/2*3/5+8/5
=13/10
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