Question

Find the equation of the tangent to the curve y=2sin^2(x) at x=pi/4

Answer 1

first find the slope of the tangent using derivatives

Answer 2

see given
y= 2*(sinx)^2
so dy/dx = 2*2(sinx)*cosx
at x=pi/4 ,y = 2*(1/sqrt(2)^2) =1
dy/dx =4* 1/(sqrt( 2) * sqrt( 2) ) =2
dy/dx = 2
so we need aline whih has a slope 2 and passes through pi/4,1
so we have eq of reqd tangent
y-1= 2*(x- pi/4)
y-1= 2*(x-22/28) and so on [using pi=22/7]