使用python适合cos模型功能

Question

我应该绘制强度与（cos（theta））** 2函数的关系图（对于Marlus Law）  但是我很固执地应该如何绘制模型函数

请告诉我以下内容是否正确：

def cos (x,a, b):
 return a*(np.cos(x))**2+ b

#Load the data
ex1_data= np.loadtxt("1 data2.txt", skiprows=2)
angle1 = [row[0] for row in ex1_data]
intensity = [row[1] for row in ex1_data]
Intensity=np.array(intensity)
cos1=np.cos(angle1)
cos1_square=(cos1)**2

percision_error_I =  Intensity * 0.0025 
accuracy_error_I = 0.001

erry = []
for i in range(len(percision_error_I)):
    erry.append(max(percision_error_I[i], accuracy_error_I))
#find optimized function and cov of x and y
p_opt , p_cov = curve_fit ( cos ,
                           cos1_square , Intensity , p0,
                           erry , True )
print(erry)


a_opt = p_opt[0]#find parameter for a
b_opt = p_opt[1]#find parameter for b

print("p_cov is ", p_cov)
print("a_opt is ", a_opt,"and b_opt is ", b_opt)

#plot model function
plt.plot(cos1_square, cos(cos1_square, a_opt,b_opt ), 'r-',label="cos1_square")
#plot error bar 
plt.errorbar(cos1_square, Intensity,np.array(erry),
             linestyle="",marker='+',label="error bar",lw=.7)
plt.legend()#show legend
#label current and voltage
plt.ylabel(' Intensity ')
plt.xlabel('cos1_square')
#provide a title for graph
plt.title("Voltage vs. cos1_square")
plt.show()

我收集了强度和角度的数据，但是我不知道如何绘制模型函数或误差线...请帮助