sin（x－3兀／4）cos（x－兀／4）＝－1／4,求cos4x

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