Question

Find the equation of the tangent to the curve `x=sin3t ,y=cos2t` at `t=pi/4dot`

Answer 1

`x_1=sin(3/4pi),x_1=1/G`
`y_1=cos(pi/2)=y_1=0`
`dy/dx=(dy/dt)/(dx/dx)=(-2sin2t)/(3cos3t)=(-2sin(3/4pi)/(3cos(3/4pi)))`
`=(-2)/(3*(1/sqrt2))`
`m=(2sqrt2)/3`
`y-0=(2sqrt2)/3(x-1/sqrt2)`
`y=(2sqrt2x)/3-2/3`
`(2sqrt2)/3x-y=2/3`.