函数y=3sin2x的最大值和最小值

对于这一类三角函数题,可以引入辅助角φ：y=5(sin2x*(3/5)+cos2x*(4/5).令cosφ=3/5,sinφ=4/5,tanφ=4/3.=5(sin2xcosφ+cos2xsinφ.=

y=√3cos²x+1/2sin2x=√3/2(cos2x+1)+1/2sin2x=√3/2cos2x+1/2sin2x+√3/2=sin(π/3)cos2x+cos(π/3)sin2x+√

y=2sin2x+2cosx-3=-2cos2x+2cosx-1=-2（cosx-12）2-12≤-12．∴函数y=2sin2x+2cosx-3的最大值是−12．故选：C．

f(x)=sin2x-√3cos2x=2(1/2*sin2x-√3/2cos2x)=2(sin2xcosπ/3-cos2xsinπ/3)=2sin(2x-π/3)最小正周期T＝2π/2=π因为-1

y=sin2x-2cos²x=sin2x-cos2x-1=根号2sin（2x+π/4）-1所以最大值为（根号2-1）,最小值为（-根号2-1）

y=2sin2x+2cosx-3中的sin2x是sinx的平方?y=2(sinx)^2+2cosx-3=-2(cosx)^2+2cosx-1=-2(cosx-1/2)^2-1/2当cosx=1/2时,

解y=sin2x-2sin^x=sin2x+cos2x-1=sin（2x+π/4）-1最小正周期T=2π/ω=π最大值=0此时x=kπ+π/8,k属于整数

y=2(sinx)^2+2cosx-3=2-2(cosx)^2+2cosx-3=-2(cosx)^2+2cosx-1=-2(cosx-1/2)^2-1/2当cosx=1/2时,函数有最大值1/2,此时

f(x)=cos2x+√3sin2x=2sin(2x+π/6)最小正周期为2π/2=π最大值为2单调增区间2x+π/6∈[2kπ-π/2,2kπ+π/2]x∈[kπ-π/3,kπ+π/6]所以单调增区

y=2(sin2xcosπ/3+cos2xsinπ/3)=2sin(2x+π/3)所以最大值=2,最小值=-2周期=2π/2=π最大值时,2x+π/3=2kπ+π/22x=2kπ+π/6x=kπ+π/

函数y=3/2sin2x-1的最大值是__1/2____；周期是____π______-1≤sin2x≤1;最大值=3/2-1=1/2;周期=2π/2=π;

sin2x=-1时,(1)-3*（-1）+a=8,a=5（2）a=5时,y=-3sin2x+5的最小值为：-3+5=2（3）当sin2x=-1时,取得最大值为8,这时2X=-π/2+2Kπ,X=-π/

y=√3cos²x+1/2sin2x=√3/2(cos2x+1)+1/2sin2x=√3/2cos2x+1/2sin2x+√3/2=sin(π/3)cos2x+cos(π/3)sin2x+√

y=cos2x-(√3)sin2x=-2[(√3/2)sin2x-(1/2)cos2x]=-2sin[2x-(π/6)].即y=-2sin[2x-(π/6)].∴该函数的最大值为2.

最大值是（根号2）－1

cos^2x=cos2x+1所以原式等于二分之根号三倍的（cos2x+1)+1/2sin2x化简得sin(2x+60)+二分之根号三最大值就是1+二分之根号三

y=2sin(2x-π/3)最大值为2最小值为-2T=2π/2=π

y=√3sin2x+cos2x=2(√3/2*sin2x+1/2*cos2x)=2(sin2xcosπ/6+cos2xsinπ/6)=2sin(2x+π/6)最大值=2此时2x+π/6=π/2+2kπ

解y=√3cos2x+sin2x+1=2(√3/2cos2x+1/2sin2x)+1=2(sin2xcosπ/3+cos2xsinπ/3)+1=2sin(2x+π/3)+1∵sin(2x+π/3)∈[

y=√3sin2x+cos2x=2(sin2xcosπ/6+cos2xsinπ/6)=2sin(2x+π/6)所以,y=2sin(2x+π/6)的最大值是2.