Question

How do you factor this expression as a single trigonometric function?

cos x - sin²x - 1

Answer 1

\[\Large\rm \cos x - \color{red}{\sin^2x} - 1\]You can do some fun stuff with this middle term, getting it in terms of cosine.

Answer 2

Remember your Pythagorean Identity involving sine and cosine?

Answer 3

I know sin²u + cos²u = 1. So sin²u = 1 - cos²u. Is that where you're going with this?

Answer 4

Mmm yes good.
Let's write our middle term in brackets like this,
\[\Large\rm \cos x - (\color{red}{\sin^2x}) - 1\]It will make it easier to substitute in this way.

Answer 5

\[\Large\rm \cos x - (\color{red}{1-\cos^2x}) - 1\]So that identity brings us here, yes?

Answer 6

Alright, I get that. I'm not sure where to go from there.

Answer 7

The reason I put brackets is because we need to distribute the negative to each term.\[\Large\rm \cos x -1+\cos^2x - 1\]

Answer 8

Combining like-terms,\[\Large\rm \cos^2x+\cos x-2\]

Answer 9

If you make a substitution something in your brain might click :)\[\Large\rm u=\cos x\]Giving us,\[\Large\rm u^2+u-2\]Any ideas? ^^
Stare at it for just a sec!!

Answer 10

Can we factor it? (u - 1)(u +2)

Answer 11

Mmm yes very good \c:/
From there just plug your cosx back in for u.

Answer 12

so (cosx - 1)(cosx + 2) is the final answer?

Answer 13

`factor this expression as a single trigonometric function`
We have our factors, and it's in terms of a single trig function (cosine)

Yes! Good job :)

Answer 14

Thank you so much!