Question

How do you find an equation of the tangent line to the curve at the given point `y=cosx` and `x=pi/4`?

Answer 1

`y=(-xsqrt2)/2+(pi+2sqrt2)/4`

Explanation:

To find the equation of the tangent line to the curve `y=cos(x)` at `x=pi/4`, start by taking the derivative of `y`.

`y'=-sin(x)`

Now plug in your value for `x` into `y'`:

`-sin(pi/4)=-1/sqrt(2)`
`=>-sqrt2/2` (This is equivalent to the above answer. I multiplied both the numerator and denominator by `sqrt2` to get the radical out of the denominator. This is not necessary).

This is the slope of the tangent line at `x=pi/4`.

To find the equation of the tangent line, we need a value for `y`. Simply plug your `x` value into the original equation for `y`.

`y=cos(pi/4)`
`y=1/sqrt2=sqrt2/2`

Now use point slope form to find the equation of the tangent line:

`y-y_0=m(x-x_0)`

Where `y_0=sqrt2/2`, `m=(-sqrt2)/2` and `x_0=pi/4`.

This gives us:

`y-sqrt2/2=(-sqrt2)/2(x-pi/4)`

Simplifying,

`y=(-xsqrt2)/2+(pi+2sqrt2)/4`

Hope that helps!