sin（x－3兀／4）cos（x－兀／4）＝－1／4,求cos4x

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sin（x－3兀／4）cos（x－兀／4）＝－1／4,求cos4xsin（x－3兀／4）cos（x－兀／4）＝－1／4,求cos4xsin（x－3兀／4）cos（x－兀／4）＝－1／4,求cos4xs

sin（x－3兀／4）cos（x－兀／4）＝－1／4,求cos4x
sin（x－3兀／4）cos（x－兀／4）＝－1／4,求cos4x

sin（x－3兀／4）cos（x－兀／4）＝－1／4,求cos4x
sin(x－3π/4) cos(x - π/4) = - 1/4,求cos4x
将要用到：2sinAcosB = sin(A+B) + sin(A-B)
和 cos2α = cos²α - sin²α = 1 - 2sin²α
sin(x－3π/4) cos(x - π/4) = 1/2 [sin(2x -π) + sin(-π/2)] = -1/2 [sin2x + 1] = -1/4
===> sin2x = -1/2
cos4x = 1 - 2 sin²2x = 1 - 2 *(1/4) = 1 - 1/2 = 1/2
这样做最简洁的,当然楼上几位的方法也值得参考!

化简出来 (sinx+cosx)^2=1/2 1+2sinxcosx=1/2 sin2x=-1/2 cos4x=1-2(sin2x)^2=1/2

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原式=[sinx*cos(3π/4) - cosx*sin(3π/4)]*[cosx*cos(π/4) + sinx*sin(π/4)]
=[(-√2/2)sinx - (√2/2)cosx]*[(√2/2)cosx + (√2/2)sinx]
=-[(√2/2)sinx + (√2/2)cosx]*[(√2/2)cosx + (√2/2)sinx]
=-(√2/2)^2*(cosx + sinx)^2
=-(1/2)[(cosx)^2 + (sinx)^2 + 2ainxcosx]
=-(1/2)(1 + sin2x) = -1/4
则：1 + sin2x = 1/2
sin2x=-1/2
cos4x =1 - 2*(sin2x)^2 =1 - 2*(-1/2)^2 =1 - (1/2) =1/2

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sin（x－3兀／4）cos（x－兀／4）＝－1／4
sin(x+兀/4-兀)cos（x－兀／4）＝－1／4
-sin(x+兀/4)cos（x－兀／4）＝－1／4
sin(x+兀/4)cos（x－兀／4）＝1／4
(sinxcos兀／4+sin兀／4cosx)(cosxcos兀／4+sinxsin兀／4)=1/4
√2/2(sinx+cosx)*√2/2(sinx+cosx)=1/4
1/2(sinx+cosx)^2=1/4
(sinx+cosx)^2=1/2
1+2sinxcosx=1/2
sin2x=-1/2 (2sinxcosx=sin2x)
(sin2x)^2=1/4
所以cos4x=1-2(sin2x)^2=1-2*(1/4)=1/2

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sin(x－3兀/4)cos(x－兀/4)＝(1/2){sin[(x－3兀/4)+(x－兀/4)]+sin[(x－3兀/4)-(x－兀/4)]}=-1/4
sin(2x-兀)+sin(-兀/2)=-1/2,-sin(兀-2x)-1=-1/2,sin2x=-1/2,cos4x=1-2(sin2x)^2=1-2*(1/4)=1/2

cos 2x /sin （x －兀/4）＝－根号2/2,则cos x ＋sin x 的值如题 sin（x－3兀／4）cos（x－兀／4）＝－1／4,求cos4x 三角化简sin(x-5兀)cos[(-兀/2)-x]cos(8兀-x)/sin[x-(3兀/2)]sin(-x-4兀) 证明sin^2(x)+cos^2(x+30)+sin(x)cos(x+30)=3/4 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x f(x)=cos(2x-兀/3）+2sin(x-兀/4)sin(x+兀/4)怎么化简化简2sin^2[(π/4)+x]+根号3（sin^x-cos^x）-1 已知函数为f（x）=cos（2x－π╱3）＋2sin（x-π/4）sin（x+π/4） 2(sin^6x+cos^6)-3(sin^4x+cos^4x) 怎样判断被积函数R（sin x,cos x）是关于sin x 的奇函数?RT,例(sin x cos^3 x）/（1+sin^4 x）是关于cos x 的奇函数,为什么? sin(5x)-sin(3x)= 根号2cos(4x)1.sin(5x)-sin(3x)= （根号2）cos(4x)解方程2.sin（x）+sin(2x)+sin(3x)=1+cos(x)+cos(2x) (1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x 化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x 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- 若x=兀/12 sin^4-cos^4= cos（兀／4-x）为什么=sin（兀／4 x）已知sin(x-兀/4)=1/3,则cos(兀/4 +x)=? 若sin(兀/3-x)=1/4,则cos(兀/3+2x)等于多少