Finding $\sin(4a)$ if we know $\cos a$ 
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*I need to show that $\sin 4a = 0$   if $\cos a = 0$.
I am not sure how to do this really.
I know I can take $\sin^2 x + \cos^2 x = 1$ but I don't think that helps.

*I was also suppose to find the period of $\sin 2x + \cos 5x$, I know the answer because I graphed it but is there a really easy formula or way to look at it and see the answer? I came up with $2\pi$ because that is where $10\pi/5$ and $2\pi$ will meet. ($\pi/5$ and $\pi$)
EDIT: I think I just got it actually, if $\cos a$ is 0 that means $a$ is a multiple of 90 degrees so 4 times that is 360, whose is sin 0.
 A: What other identities have you tried? Perhaps the double angle identity will be helpful here:    $ \sin(2a) = 2\sin(a)\cos(a) $ (This is of course for the first question, as Ben Alpert said, you should consider posting the second part as a separate question.)
A: You got it in your edit. Just be careful that $\cos x=0$ means that $x$ is an odd multiple of $90^\circ$ (the cosine of just any integer multiple of $90^\circ$ need not be $0$) and so $4x$ is a multiple of $360^\circ$ and so $\sin x=0$.
A: Starting with the addition identities:
$$\sin (a+b)=\sin a\cdot \cos b+\cos a\cdot \sin b,$$
$$\cos (a+b)=\cos a\cdot \cos b-\sin a\cdot \sin b,$$
we get (setting $a=b$)
$$\sin (2a)=2\sin a\cdot \cos a,$$
$$\cos (2a)=\cos ^{2}a-\sin ^{2}a.$$
Hence
$$\sin (4a)=2\sin (2a)\cdot \cos (2a)=2\cdot 2\sin a\cdot \cos a\left( \cos
^{2}a-\sin ^{2}a\right). $$
If $\cos a=0$, then 
$$2\cdot 2\sin a\cdot \cos a\left( \cos ^{2}a-\sin^{2}a\right) =0,$$
and so is $\sin (4a)=0$.
This is not the minimal calculation, but it is very automatic.
A: For your second question...
In general, the graph of $y=f(2x)$ looks like the graph of $y=f(x)$, except that it is compressed horizontally around the $y$-axis by a factor of $2$ (the function "goes through" the values of $x$ twice as fast; so $f$ takes all values it used to take between, say, $0$ and $2$, but does it between $0$ and $1$; so what used to "take" all the way to $x=2$ to graph now needs to be graphed between $0$ and $1$, hence "twice as fast").
Likewise, the graph of $y=f(3x)$ is like the graph of $y=f(x)$, compressed by a factor of $3$; $y=f(4x)$ is like the graph of $y=f(x)$ compressed by a factor of $4$, etc. a
That means that the graph of $y=sin(2x)$ will look just like the graph of $y=\sin(x)$, but compressed by a factor of $2$. Since $y=\sin(x)$ takes $2\pi$ to do a full cycle, then $y=\sin(2x)$ will take half as long (will do the cycle "twice as fast"). So the period is $\pi$ instead of $2\pi$.
Similarly, the graph of $y=\cos(5x)$ is like the graph of $y=\cos(x)$, but compresssed by a factor of $5$; it finishes a cycle five times as fast as the regular cosine does; so it does a cycle every $\frac{2\pi}{5}$.
If you have one function repeating every $\pi$, and one function repeating every $\frac{2\pi}{5}$, then they will "sync back up" at $2\pi$, because that is the smallest integer multiple of both $\pi$ and $\frac{2\pi}{5}$, as you surmise.
Alternatively: if $y=g(x)$ is periodic with period $P$, that means that $g(x+P)=g(x)$ for all $x$. What is the period of $y=g(2x)$? Well, if you instead of $x$ you plug in $x+\frac{P}{2}$, you get:
$$y = g\left( 2\left(x+\frac{P}{2}\right)\right) = g(2x + P) = g(2x)$$
so $\frac{P}{2}$ is a period for $y=g(2x)$. Generally, if $y=g(x)$ has period $P$, then $y=g(kx)$ has period $\frac{P}{k}$ (can you see why?). 
A: First question - haven't the rep for a comment. Apply the double angle formula twice for sin, and you get an expression sin(4a) = 4cos(a)sin(a)cos(2a) the first factor of which is zero - you don't need to expand the cos(2a) bit.
