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

Let x = 2sintheta - sin2theta and y = 2costheta - cos2theta find (d^2 y)/(dx^2) at theta= pi

Let x = 2sintheta - sin2theta and y = 2costheta - cos2theta find (d^2 y)/(dx^2) at theta= pi

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= a(costheta + sintheta), y= a(sintheta + costheta) , then (d^2y)/(dx^2) at theta = pi/4 .............

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 =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=2cos theta -cos2 theta and y=2sin theta -sin2 theta , find (d^2y)/(dx^2) at theta=pi/2 .

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

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(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) .