Now flip your original straight line about the x axis. You now have a negative angle: -theta, which is in the fourth quadrant. The adjacent side is still positive, because x is always positive in the fourth quadrant, and the hypotenuse is always positive. Therefore, the cosine function will still be positive. In addition, its value will be the same as before.

You can repeat this argument in the second and third quadrants, where the cosine function is negative (since x is negative there), to convince yourself that, indeed, the cosine function is even.

A similar argument will show you, by the definition of the sine function, that it is odd.

Alternatively, you can show the result from the Taylor series expansions of the two functions. The cosine function has only even powers of x in its expansion, and the sine function has only odd powers of x in its expansion. The result follows immediately from that.

Does this make sense?

It's pretty easy to see that with \(\displaystyle \cos{(x)}\) everything to the left of the \(\displaystyle y\) axis is a reflection of everything to the right of the \(\displaystyle y\) axis.

So \(\displaystyle f(-x) = f(x)\) which means \(\displaystyle \cos(-x) = \cos(x)\).

Welcome to Math Help Forum!

As soon as you start to look at the sine and cosine of angles outside the range \(\displaystyle 0^o\) to \(\displaystyle 90^o\), you need a definition which doesn't rely on the ratios of the sizes of the sides of a right-angled triangle.Also,

Why is sin(-x) = -sin(x)?

I'm sure I learned this all before, but I've forgotten.

thanks!

Instead, you need to look at the Unit Circle.

If you study the Wikipedia page above, you'll see that the cosine and sine of the angle \(\displaystyle \displaystyle t\) are

A moment's thought will convince you that if the point moves

So \(\displaystyle \cos (-t) = \cos (t)\).

But its \(\displaystyle \displaystyle y\)-coordinate is \(\displaystyle \displaystyle -1\) times what it was before.

So \(\displaystyle \sin(-t) = \sin(t)\)

OK now?Grandad