Question

Question #65c6d

Answer 1

Just like you normally would, just remember that trigonometric equations can have an infinite amount of solutions.

Explanation:

Critical points of trig functions really aren't that special. You just find the derivative and set it equal to `0`.

The only slightly tricky bit comes when you have to figure out when the functions equal to `0`, because they are periodic.

Let's take `sin(x)` as an example. The derivative is `cos(x)`, so if we set that equal to `0` we can find the critical points:

`cos(x)=0`

There are two solutions for this that might spring to mind, `x=pi/2` and `x=(3pi)/2`. While these certainly are solutions, they are not the only ones. Because the cosine function is periodic and repeats every `2pi`, we can add `2pik` where `k in ZZ` to get all the points where it repeats.

This means our critical points are `x=pi/2+2pik` and `x=(3pi)/2+2pik` where `k in ZZ`.

You can also simplify this to just `x=pi/2+pik` since `pi/2` and `(3pi)/2` only differ by `pi`.

Inverse trigonometric functions
You can work out the critical points to the inverse trig functions just like regular too, but you'll find that none of them have any critical points.

You'll notice that all of the derivatives of the inverse trig functions only have stuff relating to `x` in the denominator and just a `1` in the numerator, and therefor they can never equal `0`.

Let's look at `sin^-1(x)` as an example:

`d/dx(sin^-1(x))=1/sqrt(1-x^2)`

`1/sqrt(1-x^2)=0`

This above equation has no solutions, because the numerator would have to equal `0`, and it can't, since it is always `1`.