将Y=4cos^4x+4sin^4x-3化为y=Asin(wx+@)的形式,其结果为?,当Y＜0时,X的取值范围是?题目中＠为一个希腊字母,恕愚人才疏学浅,不会打出来．．．．拜托了!

将f(x)=2sin^4 x+2cos^4 x+cos^2 x-3化简化成y=Asin(wx+b)+c形式 已知函数f(x)=根号3sin2x+2sin(派/4+x)cos(派/4+x)将f(x)化为y=sin 化简y=sin^4x-cos^4x, y=sin^4x+cos^4x的值域 证明sin^2(x)+cos^2(x+30)+sin(x)cos(x+30)=3/4 函数y=cos^4x-sin^4x的最小正周期y=(cos^2x+sin^2x)(cos^2x-sin^2x)为什么等于cos^2x-sin^2x=cos2x谢谢 函数y=sin(x)cos(x+ π/4)+cos(x)sin(x+π/4)的最小正周期是 y=4√2sin x cos x+cos 2x 的值域是 6-4x-x^2=5/|sin(y/x)cos(y/x)|的解 求函数y=7-4sin(x)cos(x)+4cos^2(x)-4cos^4(x)值 y=2cos(x/4+ π/3)-3sin(x- π/4)y=sin(4x- π/3)-cos(3x+π/2) 求y=sin^5x ,y=cos(4x+3) 的导数, y=sin^2x-3sinxcosx+4cos^2x是否等于y=sin^2x+cos^2x+3cos^2x-3/2*sin2x?y=sin^2x-3sinxcosx+4cos^2xy=sin^2x+cos^2x+3cos^2x-3/2*sin2x=1+3cos^2x-3/2*sin2x=1+3（1-sin^2x）-3/2*sin2x=-3sin^2x-3/2*sin2x+4 如何将y=(sin^2(x/4)+cos^2(x/4))^2-2sin^2(x/4)cos^2(x/4)变形到(3/4)+(1/4)cosx? y=sin²x*cos^4 的最大值 已知点p(x,y)为曲线C:{x=3sinθ 4cosθ y=4sinθ-3cosθ上动点若不等式x m 将y=4sin^4x+4cos^4x-3转化为y=Asin(wx+φ)的形式 y=sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x 这一步是怎么来的?