sin4x+cos2x的最小周期

廉泰征2247化简sin^4x+cos^2x -
解秆廖19685511315 ______ (7+cos4x)/8 主要利用了降幂公式 sin^2x=(1-cos2x)/2 cos^2x=(1+cos2x)/2 你试试看 不会再找我

廉泰征2247已知函数f(x)=sin^4x+cos^2x+1/4sin2xcos2x(x∈R)则f(x)化简的到式子是? -
解秆廖19685511315 ______[答案] f(x)=sin^4x+cos^2x+1/4sin2xcos2x =(1/4)[1-cos2x]^2+1/2(cos2x+1)+(1/8)sin4x =(1/4)-(1/2)cos2x+(1/4)(cos2x)^2+(1/2)cos2x+1/2+(1/8)sin4x =3/4+(1/8)(1+cos4x)+(1/8)sin4x =7/8+(1/8)(cos4x+sin4x) =(√2/8)sin(4x+π/4)+7/8 其最小正周期T=2π/4=π/2

廉泰征2247函数f(x)=sin^4x+cos^2x的最小正周期 -
解秆廖19685511315 ______[答案] f(x)=sin^4x+cos^2x =1-sin^2x+sin^4x =sin^2x(sin^2-1)+1 =-sin^2xcos^2x+1 =-(1/2sin2x)^2+1 =-1/8(1-cos4x)+1 =7/8+1/8cos4x 最小正周期为1/2π

廉泰征2247解不等式:sin4x+cos4x*cot2x>1 求其过程 -
解秆廖19685511315 ______ 解答:sin4x+cos4x*cos2x/sin2x>1(sin4xsin2x+cos4xcos2x)/sin2x>1 cos(2x)/sin(2x)>1 cot2x>1 所以 kπ<2x<kπ+π/4 所以 kπ/2<x<kπ/2+π/8,k∈Z

廉泰征2247y=sin4x+cos2x, dy=? -
解秆廖19685511315 ______ 因为y'=(sin4x+cos2x)' =4cos4x-2sin2x 所以 dy=(4cos4x-2sin2x)dx 不懂可追问 满意请采纳 谢谢

廉泰征2247化简:sin4x*cos2x*cosx/(1+cos4x)/(1+cos2x)/(1+cosx) -
解秆廖19685511315 ______[答案] 1+cos2x=2cos² 分母为cos²2x*cos²x*(1+cosx) 分子为4sinx*cos²2x*cos²x 化简为4sinx/(1+cosx)=4tan(x/2)

廉泰征2247数学sin4x - cos4x -
解秆廖19685511315 ______ sin^4 x - cos^4 x = (sin^2 x + cos^2 x)(sin^2 x - cos^2 x) = 1*(-cos (2x)) = - cos (2x) 周期 = pai

廉泰征2247解不等式:sin4x+cos4x·cot2x>1 -
解秆廖19685511315 ______[答案] =2.sin2x.cos2x+[(cos2x)^2-(sin2x)^2].(cos2x/sin2x)=2.sin2x.cos2x+(cos2x)^3/sin2x-sin2x.cos2x=sin2x.cos2x+(cos2x)^3/sin2x=[(sin2x)^2.cos2x+(cos2x)^3]/sin2x=cos2x/sin2x>1得到cot2x>1即n*pi

廉泰征2247已知cos2x=五分之三,求sin4x+cos4x的值 -
解秆廖19685511315 ______ 解:sin4x=2sin2xcos2x 用sin2x^+cos2x^2=1解即可 sinx^4=(sinx^2)^2=(1-cos2x/2)^2 然后相加即可 很高兴为您解答,祝你学习进步!【学习宝典】团队为您答题.有不明白的可以追问!如果您认可我的回答.请点击下面的【选为满意回答】按钮,谢谢!

廉泰征2247求(sinx)^4(cosx)^2的不定积分 -
解秆廖19685511315 ______[答案] ∵(sinx)^4(cosx)^2 =(1-cos2x)^2(1+cos2x)/8 =[1-(cos2x)^2](1-cos2x)/8 =(sin2x)^2(1-cos2x)/8 =[1-(cos4x)]/16-(sin2x)^2(cos2x)/8 ∴原积分=∫[1-(cos4x)]/16*dx-∫(sin2x)^2(cos2x)/8*dx =x/16-(sin4x)]/64-1/16*∫(sin2x)^2(dsin2x) =x/16-(sin4x)]/64-(sin2x)^3/48...