How do you find an equation to the tangent line to the curve #y = 2 sin^2x# at #x = Pi/4#?

Question

Luke Alvarez · Accepted Answer

#y=2x+1-pi/2# The tangent line could be represented as : #y=mx+c# where #color(red)m# is the gradient of that tangent, and #m=((dy)/(dx))_(x=pi/4)# #color(red)c# is the #y#-intercept First find #m# : #y=2sin^2x# #(dy)/(dx)=2*2sinx*cosx=2sin(2x)# Since, #m=((dy)/(dx))_(x=pi/4)# #=>m=2sin(2*pi/4)=2*1=color(red)2# The equation of tangent is now : #y=2x+c# To find #c# we plug in the values of #x# and #y# at the point where the tangent touches the curve(#y=2sin^2x#) The tangent is at the point, where #x=pi/4# The corresponding #y# is found by putting #x=pi/4# inside #y=2sin^2x# #=>y=2sin^2(pi/4)=2*(1/sqrt2)^2=2*1/2=1# Plug those into #y=2x+c# #=>1=2*pi/4+c# Where looking for #c# #=>c=color(red)(1-pi/2)# Hence , tangent is : #color(blue)(y=2x+1-pi/2)#

Paisley Johnson · Answer

undefined To find the equation of the tangent line to the curve y = 2 sin^2x at x = Pi/4, we need to find the slope of the tangent line and a point on the line.

First, find the derivative of the function y = 2 sin^2x using the chain rule: 
dy/dx = 2 * 2 sin(x) * cos(x) = 4 sin(x) cos(x)

Evaluate the derivative at x = Pi/4: 
dy/dx = 4 sin(Pi/4) cos(Pi/4) = 4 * (1/sqrt(2)) * (1/sqrt(2)) = 4/2 = 2

The slope of the tangent line is 2.

Next, find the y-coordinate of the point on the curve at x = Pi/4: 
y = 2 sin^2(Pi/4) = 2 * (1/2)^2 = 2 * 1/4 = 1/2

The point on the curve is (Pi/4, 1/2).

Using the point-slope form of a line, the equation of the tangent line is: 
y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the line.

Plugging in the values, we get: 
y - 1/2 = 2(x - Pi/4)

Simplifying the equation gives the equation of the tangent line: 
y = 2x - Pi/2 + 1/2