If `x=2sin theta-sin2 theta` and `y=2cos theta-cos2 theta`,`theta in[0,2 pi],` then `(d^(2)y)/(dx^(2))` at `theta=pi` is

If x=cos theta-cos2 theta,y=sin theta-sin2 theta then (dy)/(dx) is

if x =2 cos theta - cos 2 theta y = 2 sin theta - sin 2 theta Find (d^(2)y)/(dx^(2)) at theta = pi/2

If x=sin theta|sin theta|,y=cos theta|cos theta|, where 99(pi)/(2)<=theta<=50 pi, then

If x=2cos theta -cos 2theta ,y=2sin theta -sin 2theta ,then (dy)/(dx) =

If x=sin theta-theta cos theta , y=cos theta+theta sin theta then (d^(2)y)/dx ^2=

If x= 2 sintheta - sin2theta and y= 2 costheta - cos 2theta then (d^2y)/(dx^2) " at " theta =pi is :

If (cos theta+cos2 theta)+sin2 theta(1+2cos theta)=2sin theta,theta in[-pi,pi] then

If x=2 cos theta- cos 2 theta and y=2 sin theta - sin 2 theta, then (dy)/(dx) =

If x=a(cos 2 theta+2 theta sin 2 theta) " and" y=a(sin 2 theta - 2 theta cos 2 theta) , find (d^(2)y)/(dx^(2)) " at" theta =(pi)/(8) .

If x=a (theta-sin theta) and y=a(1-cos theta) , find (d^(2)y)/(dx^(2)) at theta=pi.