UC知道一道三角函数题~~~比较急~~~+分
来源：百度知道 编辑：UC知道 时间：2024/05/16 18:01:28
g(x)=1-cos^2 x ,,where x is in radian
1--- the graph of function g can also be discribed by
h(x)=a cos(b(x-派/2))+d
问a,b,d的值
之后
2--- rewrite the function h as a sine function in the form
j(x)= a sin(b(x-c))+d
1--- the graph of function g can also be discribed by
h(x)=a cos(b(x-派/2))+d
问a,b,d的值
之后
2--- rewrite the function h as a sine function in the form
j(x)= a sin(b(x-c))+d
g(x)=1-cos^2x
可以表示为:
h(x)=acos(b(x-pai/2))+d
g(x)=1-cos^2x
=1-(1+cos2x)/2
=-1/2cos2x+1/2
=1/2cos(2x-pai)+1/2
=1/2cos(2(x-pai/2))+1/2
对比有:
a=1/2,b=2,c=1/2
2.
h(x)=1/2cos(2(x-pai/2)+1/2
=-1/2cos2x+1/2
==1/2sin(2x+pai/2)+1/2
=-1/2sin(2(x+pai/4))+1/2
其中
a=-1/2,b=2,c=-pai/4,d=1/2
能不能用中文提问