f (x)=cos^4(x)-sin^4(x)+5的最小正周期是?
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/09 08:33:40
![f (x)=cos^4(x)-sin^4(x)+5的最小正周期是?](/uploads/image/z/5347131-51-1.jpg?t=f+%28x%29%3Dcos%5E4%28x%29-sin%5E4%28x%29%2B5%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E6%98%AF%3F)
xJ0_҅ + x^z,"td0DAcؗ`C\/z5K=Р'Gf}usc>y98EA$٢khxE7
f (x)=cos^4(x)-sin^4(x)+5的最小正周期是?
f (x)=cos^4(x)-sin^4(x)+5的最小正周期是?
f (x)=cos^4(x)-sin^4(x)+5的最小正周期是?
f(x)=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2=1 -[sin(2x)]^2/2=1-[1-cos(4x)]/4=cos(4x)/4 +3/42π/4=π/2函数的最小正周期为π/2
先写
原式=[cos^2(x)]^2-[sin^2(x)]^2+5
=[1/2(cos2x+1)]^2-[1/2(1-cos2x)]^2+5
=1/4[(cos2x+1)^2-(cos2x-1)^2]+5
=1/4*4cos2x)+5
=cos2x+5
最快答案好像带错公式了,周期是π=[1/2...
全部展开
先写
原式=[cos^2(x)]^2-[sin^2(x)]^2+5
=[1/2(cos2x+1)]^2-[1/2(1-cos2x)]^2+5
=1/4[(cos2x+1)^2-(cos2x-1)^2]+5
=1/4*4cos2x)+5
=cos2x+5
最快答案好像带错公式了,周期是π
收起
f(x)=(6sin^4x-7sin^2x+2)/(sin^2x-cos^2x)
f(x)=sin(x+π/4)图像怎样变成F(x)=cos(2x)
证明∫( 0,π/2 ) (f sin x/(f sin x+f cos x) dx=π /4
已知f(x)=cos^2x/1+sin^2x求f'(π/4)求导
奇偶性f(x)=sin(x+/4)+cos(x+/4)判断函数奇偶性f(x)=sin(x+派/4)+cos(x+派/4)
f(x)=2sin^2(π/4+x)+根号3(sin^2x+cos^2x)f(x)=2sin^2(π/4+x)+根号3(sin^2x-cos^2x) 符号打错了
f(x)=(2sin^4 x)+(2cos^4 x)+(cos^2 2x) -3 化简
f(X)=sin^4x-cos^4x+cos^2x 判断奇偶性 要具体过程的.
已知f(x)=sin(π/6-x)^2-cos(π/4+x)^2+cos(π/6)cos(π/6-2x)化简f(x)
已知函数f(x)=cos^4x-2cos^22x+sin^2x的最小正周期
已知函数f(x)=2(sin^4 x+cos^4 x)+m(sin^x+cosx)^4在0=
设f(cos^2x)=sin^2x,则f'(π/4)等于,
函数f(x)=sin^4x+cos^4x的最小正周期
函数f(x)=sin^4x+2sinxcosx+cos^4x 的最大值是多少?
函数f(x)=(sin^4)x+2sinxcosx+(cos^4)x的最小值
f(x)=sin^4x+2√3sinxcosx-cos^4x化简
f(x)=cos^4 x-2sinxcosx-sin^4 x 的最小正周期
f(x)=cos^4x+2sinxcosx-sin^4x的最小正周期