sin(π/4-3x)cos(π/3-3x)- sin(π/4+3x)sin(π/3-3x)=

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sin(π/4-3x)cos(π/3-3x)-sin(π/4+3x)sin(π/3-3x)=sin(π/4-3x)cos(π/3-3x)-sin(π/4+3x)sin(π/3-3x)=sin(π/4-

sin(π/4-3x)cos(π/3-3x)- sin(π/4+3x)sin(π/3-3x)=
sin(π/4-3x)cos(π/3-3x)- sin(π/4+3x)sin(π/3-3x)=

sin(π/4-3x)cos(π/3-3x)- sin(π/4+3x)sin(π/3-3x)=
sin(π/4-3x)=sin[π/2-(π/4+3x)]
=cos(π/4+3x)
sin(π/4-3x)cos(π/3-3x)- sin(π/4+3x)sin(π/3-3x)
=cos(π/4+3x)cos(π/3-3x)- sin(π/4+3x)sin(π/3-3x)
=cos(π/4+3x+π/3-3x)=cos(π/4+π/3)
=√2/2*1/2-√2/2*√3/2
=(√2-√6)/4

化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1 化简sin(x+7π/4)+cos(x-3π/4)步骤我已经找到撒sin(x+7π/4)+cos(x-3π/4)=sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?=sin(x-π/4)+cos[-(π-x-π/4)]=sin(x-π/4)+cos(π-x-π/4)=sin(x-π/4)-cos(x+π/4)=sin(x-π/4)-cos(x+π/2- sin cos之间怎么转换?sin(π/3-2X)等于cos多少? tanx=m,则{sin(x+3π)+cos(π+x)}/{sin(-x)-cos(π+x)}=? 已知sin[a-b]cos a-cos[b-a]sin a=3/5,b是第三象限角,求sin[b+5π/4]的值第一题1/2cos x-√3/2sin x第二题√3sin x+cos x第三题√2[sin x-cos x]第四题√2cos x-√6sin x 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π) 化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π) 三角函数的计算题sin(X-2π)+sin(-x-3π)cos(X-3π)/cos(π-x)-cos(-π-x)cos(x-4π)等于多少? cos (2x-π/3)=2sinπ/4 sin(π/4-3x)cos(π/3-3x)- sin(π/4+3x)sin(π/3-3x)= cos(x+π/3)=1/4 sin(x+4π/3)=? sin(x- 3π/4)cos(x- π/4)=—1/4 求cos4x 化简:sin(π/3+4x)+cos(π/6-4x) 设f(x)=(sin^2(6π+x)+cosx-2cos^3(3π+x)-3)/2+cos^2(x-4π)-cos(-x)设f(x)=(sin^2(6π+x)+cosx-2cos^3(3π+x)-3)/2+cos^2(x-4π)-cos(-x)求f(π/3)的值 cos(x+π/3)是不是等于sin(x-π/6) cos(π/6-x)=sin(π/3+x)?为什么 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 化简:[sin(2π-2)cos(π+x)tan(π-x)]/[cos(π-x)sin(3π-x)sin...化简:[sin(2π-2)cos(π+x)tan(π-x)]/[cos(π-x)sin(3π-x)sin(-π-x)]