Question

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

Answer 1

`y=2x+1-pi/2`

Explanation:

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)`