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已知sin π 4 x 3 5

f(x)=sin(5π/6-2x)-2sin(x-π/4)cos(x+3π/4) = sin[π-(5π/6-2x)] - { sin[(x-π/4)+(x+3π/4)] + sin [(x-π/4)-(x+3π/4)] } = sin(2x-π/6) - { sin(2x+π/2) - sin(-π) = sin(2x-π/6) - cos(2x) = sin2xcosπ/6-cos2xsinπ/6-cos2x = √3/2sin2x-3/2cos...

√2/2sinx-√2/2cosx=3/5 sinx-cosx=3√2/5 平方 1-2sinx cosx=18/25 7/25=2sinx cosx sin2x=7/25

-4/3 解析: tan(θ-π/4) =-tan(π/4-θ) =-cot[π/2-(π/4-θ)] =-cot(π/4+θ) =-cos(π/4+θ)/sin(π/4+θ) =-4/3

(1)∵ sin (α- π 4 )= 3 5 , π 4 <α< 3π 4 ,∴ 0<α- π 4 < π 2 ,有 cos (α- π 4 )= 4 5 ;(4分)(2) sinα=sin (α- π 4 + π 4 ) = sin (α- π 4 )cos π 4 +cos (α- π 4 )sin π 4 = 7 2 10 ;(5分)(3)函数 y=cos (x- π 4 ) 的图象可...

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sin(x+π/4)=3/5 sin(x-π/4)=4/5 两式相加得: 2sinxcos(π/4)=7/5 同理,两式相减得: 2cosxsin(π/4)=-1/5 上两式相除得: tanx=-7

π/4

您好: 解答如下 17π/12

sin²(x+π/3)+cos²(x+π/3)=1 所以cos(x+π/3)=±√15/4 所以sinx =sin[(x+π/3)-π/3] =sin(x+π/3)cosπ/3-cos(x+π/3)sinπ/3 =(1+√45)/8或(1-√45)/8

sin(π/4-x)=sinπ/4cosx-cosπ/4sinx=3/5 (cosx-sinx)*√2/2=3/5 平方 (1/2)(sin²x+cos²x-2sinxcosx)=9/25 1-sin2x=18/25 sin2x=7/25

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