# When can you use a triangle to represent trig substitution?

I apologize if this seems like a newbie question but it's been really messing with my head.

I'm reading about using trig substitution in integration and everyone seems to simply construct a right triangle to get what the trig functions will be (in terms of the original variable).

Shouldn't that only work for cases where the angle is acute and positive? What about the cases where the angle is obtuse or where some trig functions might be negative. As far as I know, using the right triangle will always give a positive cos/sin/tan (because sides are always positive).

Can you please explain in as much detail as possible how to correctly use a right triangle in cases where the angle is obtuse or one of the trig functions is negative. Thanks a lot in advance.

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Read the basic answer off the triangle. Adjust sign as appropriate. – André Nicolas Jun 23 '12 at 21:32
No need to apologize; this is a very reasonable question to ask. – Qiaochu Yuan Jun 23 '12 at 21:34

It has to do with the "permanence of functional equations":

Any trig function $f$ (as, e.g., $\cos$ or $\tan$) can be extended in a natural way to an analytic function $f:\ \Omega_f\to{\mathbb C}$ with some domain $\Omega_f\subset{\mathbb C}$. Consider now a trig identity of the form $$\bigl(\phi(z):=\bigr)\quad p\bigl(f(z),g(z)\bigr)\equiv0\ ,$$ where $f$ and $g$ are trig functions and $p$ is a polynomial in two variables (there are also other types, involving, e.g., shifting or doubling the argument). A simple example would be the identity $$\cos^2 z+\sin^2 z-1\equiv0\ .$$ If you can prove such an identity for all $z=z_n$ for a sequence $(z_n)_{n\geq0}$ converging to a point $z_*\in\Omega:=\Omega_f\cap\Omega_g$ then by a standard theorem in complex analysis it is automatically true for all $z\in\Omega$, because the analytical function $\phi:\ \Omega\to{\mathbb C}$ has to be $\equiv0$ in this case.

In particular it is sufficient to prove such an identity for all real $z$ in the interval $\ \bigl]0,{\pi\over2}\bigr[\$.

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The reason you don't have to worry about if it is obtuse is because of the range if the inverse trig functions , ex.) the range of $\arcsin(x)$ is $[-\pi/2 ,\pi/2]$ . And normally if you use the right trig function you shouldn't get the side of a triangle to be negative when it comes to the case of trig substitution. I know this answer was pretty vague and if you need any more info just comment on it but what you should get out of it is that the inverse trig functions restrict the angles

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The range of the sin function (over the reals) is [-1, 1]. The domain is all of $\mathbb{R}$. By context presumably you mean to write the arcsin function, but even that's a bit misleading. I definitely see the point you're after here, but this seems like a really confusing way of phrasing it... – Steven Stadnicki Jul 2 '12 at 19:04
Thank you and yes I was referring to arcsin – thebmags Jul 3 '12 at 12:26