已知tanα=-1/2,求下列各式的值;sinα+2cosα
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1.(1)原式=(4tan-2)/(5+3tan)【分子分母同除以tan】
=10/14
=5/7
(2)原式=(sincos)/(sin^2+cos^2)
=tan/tan^2+1
=3/9+1
=3/10
(3)原式=sin^2+cos^2+2sincos
=1+2*3/10
=8/5
(4)貌似题目有打错了.
2.sin(π+α)=-sinα=-1/2
∴sinα=1/2
∴cosα=±根号3/2
(1)原式=cos(-a)
=cosa
=±根号3/2
(2)原式=-tana=-sina/cosa
=(-1/2)/(±根号3/2)
=±根号3/3
3.(1)
cos25π/6+cos25π/3+tan(-25π/4)
=cosπ/6+cosπ/3-tanπ/4
=根号3/2+1/2-1
=(根号3-1)/2
(2)
sin2+cos3+tan4【无特殊角,只能计算器算】
sin2≈0.9093
cos3≈-0.9900
tan4≈1.1578
加起来既得:0.0771
【打了很久,
=10/14
=5/7
(2)原式=(sincos)/(sin^2+cos^2)
=tan/tan^2+1
=3/9+1
=3/10
(3)原式=sin^2+cos^2+2sincos
=1+2*3/10
=8/5
(4)貌似题目有打错了.
2.sin(π+α)=-sinα=-1/2
∴sinα=1/2
∴cosα=±根号3/2
(1)原式=cos(-a)
=cosa
=±根号3/2
(2)原式=-tana=-sina/cosa
=(-1/2)/(±根号3/2)
=±根号3/3
3.(1)
cos25π/6+cos25π/3+tan(-25π/4)
=cosπ/6+cosπ/3-tanπ/4
=根号3/2+1/2-1
=(根号3-1)/2
(2)
sin2+cos3+tan4【无特殊角,只能计算器算】
sin2≈0.9093
cos3≈-0.9900
tan4≈1.1578
加起来既得:0.0771
【打了很久,
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