Study the parity of $\cos(\pi+x)$, $\sin(\pi-x)$ and $\sin(\frac{3\pi}{2}+x)$ I tried to apply the following properties:
to be even:
$$f(x) = f(-x)$$
to be odd:
$$-f(x) = f(-x)$$
1)
to be even:
$$\cos(\pi+x)  = \cos(-(\pi+x)) \Leftrightarrow -\cos(x) = ??$$
What is $\cos(-(\pi+x))$ equal to?
to be odd:
$$\cos(-(\pi+x)) = -\cos(\pi+x)\Leftrightarrow ?? = \cos(x)$$
Again, same problem.
2)
to be even:
$$\sin(\pi-x) = \sin(-(\pi-x)) \Leftrightarrow \sin(x) = ??$$
What is $\sin(-(\pi-x))$ equal to?
to be odd:
$$\sin(-(\pi-x)) = -\sin(\pi-x) \Leftrightarrow ?? = -\sin(x)$$
Again, same problem.
3)
I don't really know how to solve this one.
Can anyone help me solve these? Also, is there any property that I can use to solve things like $\sin(k\alpha)$, being k a constant?
 A: You're setting them up incorrectly. You need to understand function notation better: $f(-x)$ means that $\color{red}{x}$ is replaced with $\color{red}{(-x)}$, not that you negate something somewhere. For example, if
$$f(x)=\cos(\pi+\color{red}{x}),$$
then
$$f(-x)=\cos(\pi+\color{red}{(-x)})=\cos(\pi-x),$$
not $\cos(-(\pi+x))$.
Then you can continue simplifying using properties of trigonometric functions, such as formulas for sums and differences of angles. For example, there are formulas stating that $\cos(\pi+\alpha)=-\cos\alpha$ and $\cos(\pi-\alpha)=-\cos\alpha$ for any $\alpha$. So in this question we can continue as follows:
$$f(-x)=\cos(\pi+(-x))=\cos(\pi-x)=-\cos x$$
and
$$f(x)=\cos(\pi+x)=-\cos x.$$
We can see that $f(-x)=f(x)$, i.e. this is an even function.
You can handle the other ones similarly.
A: let $f_1(x) = \cos(\pi+x)$
$$f_1(-x) = \cos(\pi-x) =\cos(\pi)\cos(-x)-\sin(\pi)\sin(-x) = - \cos(-x) $$
$$= -\cos(x) =\cos(\pi+x) = f_1(x)$$
conclusion : $f_1$ is odd
it works pretty much the same for the other functions just replace $x$ by $-x$ simplificate and conclude.
A: Also, to answer your question about $\sin(kx)$, there is a way to do this.
$$\sin(2x) = 2\sin(x)\cos(x)$$
$$\cos(2x) = \cos^2(x) - \sin^2(x)$$
both of which can be proved using the sum identity ($2x = x + x$). To solve something with $k > 2$, proceed like this;
$$\sin(3x) = \sin(2x + x) = \sin(2x)\cos(x) + \cos(2x)\sin(x) = ...$$
where you can replace $\sin(2x),\cos(2x)$ with the previously mentioned identities and simplify.
