# How to simplify a complex expression involving trig functions?

Specifically (it wouldn't fit into the question title), I'm trying to figure out how to turn

$$\frac{-2v^2}{a^2} \cdot \sin({\arctan({\frac{vt}{a}}})) \cdot \cos^3(\arctan(\frac{vt}{a}))$$ into $$\frac{-2v^3 \cdot a \cdot t}{(a^2+v^2t^2)^2}$$

Where $v$ and $a$ are constants and $t$ is the independent variable.

Wolfram Alpha tells me the two expression are indeed equivalent, but I can't seem to turn the first one into the second one algebraically.

My thinking was, $\arctan({\frac{vt}{a}})$ gives me the angle $x$ such that $\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{vt}{a}$, so shouldn't $\sin(x) = vt$ and $\cos(x)=\frac{1}{a}$?

But that doesn't give me the right result, and I can see a flaw in that thinking because $\frac{\sin(x)}{\cos(x)} = \frac{vt}{a}$ just gives use the ratio of $\sin(x)$ to $\cos(x)$, and we can't deduce from that the actual values of $\sin(x)$ and $\cos(x)$.

But so, how do you turn the first expression into the second one? If WA can tell me they are equivalent, I assume there's a way to show it.

For any $y$, $$\sin^2(y)+\cos^2(y)=1$$ In particular this is true for $y=\arctan{x}$. So $$\sin^2(\arctan(x))+\cos^2(\arctan(x))=1$$ dividing by $\cos^2$ we get $$\tan^2(\arctan(x))+1=\sec^2(\arctan(x))$$ But this means $$x^2+1=\sec^2(\arctan(x))$$ See if you can use this.
• For the final touch on my hint, note that $\sin(x)\cos^3(x)=\tan(x)\cos^4(x)$. Commented Nov 24, 2015 at 0:29
• This does not seem to answer the question. If you kept going, however, you would note that $$\sqrt{x^2+1}=sec(arctan(x))=\frac 1 {cos(arctan(x))}$$, which is more useful. Also, dividing by $$sin^2$$ instead of $$cos^2$$ will give you the other trig identity needed to solve this equation. Commented Nov 24, 2015 at 0:31
Note that $sin(arctan(x))=\frac{x}{\sqrt{1+x^2}}$ and $cos(arctan(x))=\frac{1}{\sqrt{1+x^2}}$. This can be proven with $\frac 1 {cos(A)}=sec(A)=\sqrt{1+[tan(A)]^2}$ and $A=tan(x)$. You get the formulas from your trig identities, which, in this case, is derived from the Pythagorean theorem.
$sin(arctan(x))$ can be found similarly.