回答:
证明如下
说明:
注意
你如何简化f(theta)= cos ^ 2theta-sin ^ 2theta-cos2theta?
F(theta)= 0 rarrf(theta)= cos ^ 2theta-sin ^ 2theta-cos2theta = cos2theta-cos2theta = 0
Sin ^ 2theta-cos ^ 2theta = 1-2sin ^ 2theta?
“否”“几乎:”sin ^ 2(theta) - cos ^ 2(theta)= 2 sin ^ 2(theta) - 1 sin ^ 2(theta)+ cos ^ 2(theta)= 1 => sin ^ 2 (θ) - cos ^ 2(theta)= sin ^ 2(theta) - (1 - sin ^ 2(theta))= 2 sin ^ 2(theta) - 1
在θ=(pi)/ 4处,r =(sin ^ 2theta)/( - thetacos ^ 2theta)的切线斜率是多少?
斜率是m =(4 - 5pi)/(4 - 3pi)这里是对具有极坐标的切线的参考。从参考中,我们得到以下等式:dy / dx =((dr)/(d theta)sin( theta)+ rcos(theta))/((dr)/(d theta)cos(theta) - rsin(theta))我们需要计算(dr)/(d theta)但是请注意r(theta)可以是通过使用身份sin(x)/ cos(x)= tan(x)简化:r = -tan ^ 2(θ)/ theta(dr)/(dθ)=(g(theta)/(h(theta) )))'=(g'(θ)h(θ) - h'(θ)g(theta))/(h(θ))^ 2 g(θ)= - tan ^ 2(θ)g'( theta)= -2tan(theta)sec ^ 2(θ)h(θ)= theta h'(θ)= 1(dr)/(dθ)=( - 2thetatan(θ)sec ^ 2(θ)+ tan ^ 2(θ)/(θ)^ 2让我们在pi / 4秒^ 2(pi / 4)= 2 tan(pi / 4)= 1 r'(pi / 4)=(-2( pi / 4)(1)(2)+ 1)/(pi / 4)^ 2 r'(pi / 4)=( - 2(pi / 4)(1)(2)+ 1)(16 /( pi ^ 2))r'(pi / 4)=(16-16pi)/(pi ^ 2