1. ## Trig in calc!

I've been teaching myself calc this summer, and the concepts are coming fine, but I'm struggling with using trig in examples. I don't really understanding taking the derivatives of inverse and normal trig fxns and I've skipped it when doing integrals as well. What do I need to master in order to understand this stuff, specifically? The fxns and identities just don't make sense to me algrebraically. Any websites I could visit that could help? It's appreciated!

2. ## Re: Trig in calc!

I'll try to help by way of an example. Consider $y=\arcsin x$. To find the derivative of this it's best to take sines of both sides to get to a derivative you already know, specifically:

$\sin y=x$

From here, the derivative with respect to $x$ (not $y$!) is:

$\frac{dy}{dx}\cos y=1$

which becomes

$\frac{dy}{dx}=\frac{1}{\cos y}$

Hopefully these steps are fairly clear? We know from the start of the question that $y=\arcsin x$, so

$\frac{dy}{dx}=\frac{1}{\cos\left(\arcsin x\right)}$

but this is a pretty ugly answer. It can be improved by considering the trig identity:

$\sin^2 y + \cos^2 y=1$
$\therefore \cos y=\sqrt{1-\sin^2 y}$

but we know $x=\sin y$, so

$\therefore \cos y=\sqrt{1-x^2}$

and therefore

$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$

I hope that helps!

3. ## Re: Trig in calc!

I just realised I missed the wider point of your question! Though perhaps it will be helpful later.

For now I'll just focus on something that should help you with the trig identities.

If you consider the right-angled triangle with base $x$, height $y$, and hypotenuse $h$, with $\theta$ being the angle between sides $x$ and $h$, then obviously the trig functions can be expressed as

\begin{align*}\sin\theta&=&\frac{y}{h}\\ \cos\theta&=&\frac{x}{h}\\ \tan\theta&=&\frac{y}{x}\end{align*}

Given that you say you are having a hard time understanding the algebra of the identities, hopefully the following approach will help. Replace the trig functions themselves with the equations above, and go from there. For example, one of the first trig identities you will likely have come across is:

$\sin^2\theta+\cos^2\theta\equiv 1$

If you replace sine and cosine with the above equations, this becomes

$\left(\frac{y}{h}\right)^2+\left(\frac{x}{h}\right )^2=\frac{y^2}{h^2}+\frac{x^2}{h^2}=\frac{x^2+y^2} {h^2}$

But, via Pythagoras we know that $x^2+y^2=h^2$, therefore

$\frac{x^2+y^2}{h^2}=\frac{h^2}{h^2}=1$

Hopefully the reason that identity holds becomes clear from that method. Similar methods can be applied to other trig identities. If you can figure out (at least a number of) the remaining identities yourself, you'll make some progress towards understanding the whole.