已知cosx=-3/5,-3/2π

问题描述:

已知cosx=-3/5,-3/2π

数学人气:357 ℃时间:2020-05-08 02:43:41
优质解答
∵cosx=-3/5 ,-3π/2∴sinx=-4/5,tanx=4/3
sin(π/2-2x)-2cos²(π/4+x)=cos2x-[1+cos(π/2+2x)]=cos2x+sin2x-1
=cos²x-sin²x+2sinxcosx-1=9/25-16/25+24/25-1=16/25
1+tan(π/4-x)=1+[(tanπ/4-tanx)/(1+tanπ/4tanx)]
=1+[(1-4/3)/(1+4/3)]=6/7
所以原式=(16/25)/(6/7)=56/75
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∵cosx=-3/5 ,-3π/2∴sinx=-4/5,tanx=4/3
sin(π/2-2x)-2cos²(π/4+x)=cos2x-[1+cos(π/2+2x)]=cos2x+sin2x-1
=cos²x-sin²x+2sinxcosx-1=9/25-16/25+24/25-1=16/25
1+tan(π/4-x)=1+[(tanπ/4-tanx)/(1+tanπ/4tanx)]
=1+[(1-4/3)/(1+4/3)]=6/7
所以原式=(16/25)/(6/7)=56/75