# Do trigonometric reduction formulae work for every angle?

I'm doing some basic trig exercises like if $\sin(\theta)=k$ then find $\sin(\pi+\theta)$, etc. For instance, we know that $\sin(\pi+\theta) = -\sin(\theta)$ but is the angle $\theta$ required to be $0<\theta<\frac{\pi}{2}$ for those formulae to work? Or do the reduction formulae work for all angles? My exercises of this type contain conditions like $0<\theta<\frac{\pi}{2}$ or $\pi<\theta<\frac{3\pi}{2}$. Are those statements reduntant?

EDIT: By the reduction formulae I mean all formulas for simplifying trigonometric expressions like this e.g. $\cot(\frac{\pi}{2}+x)=-\tan(x)$, etc. So are all such formulas derived from identities and are therefore identities too i.e. they work for every angle, regardless of the quadrant?

• In some precise sense*, the nature of trig functions ensures that if a "reasonably well behaved" identity holds for an interval of $\theta$, it holds for all of them. "Reasonably well behaved" excludes square roots and absolute values among other things - so it applies to expressions like $\cos(\theta)^2+\sin(\theta)^2=1$ which only use addition and multiplication, but doesn't apply to $\sqrt{1-\sin(\theta)^2}=\cos(\theta)$, which holds in $[-\frac{\pi}2,\frac{\pi}2]$ but not everywhere - like at $\pi$. (*This precise sense invokes complex analysis, so might not be super accessible...) – Milo Brandt Nov 15 '15 at 17:39

Yes, in general trigonometric identities work for all angles big and small, positive or negative -- unless they involve non-smooth operations like absolute values or piecewise definitions.

Conditions like $0<\theta<\pi/2$ can usefully appear in problems. They don't usually impact the validity of the identifies you use, but can serve to select a particular branch of the inverse trigonometric functions. For example if you're looking for $\theta$ such that $\sin\theta = 1/2$, then knowing $0<\theta<\pi/2$ will allow you to discard $\theta=\frac23\pi$, $\theta=\frac73\pi$, and so forth, leaving $\theta=\frac13\pi$ as the only solution.

$$\sin(\pi+\theta)=\sin\pi\cos\theta+\cos\pi\sin\theta=0\cdot\cos\theta+(-1)\sin\theta=-\sin\theta$$

So, this is an identity

• By the reduction formulae I mean ALL formulas for simplifying trigonometric expressions like this e.g. $cot(\frac{\pi}{2}+x)=-tan(x)$, etc. So are all such formulas derived from identities and are therefore identities too i.e. they work for every angle, regardles of the quadrant? – Richard Smith Nov 15 '15 at 15:12
• @RichardSmith: $\cot(\frac{\pi}2+x) = -\tan(x)$ has problems when $x=\frac{\pi}2, \frac{3\pi}2$ etc., for the same reason that $\cot(\pi)$ and $\tan(\frac{\pi}2)$ have problems on their own: they are not well defined. But where they are well defined, this identity holds – Henry Nov 15 '15 at 22:33

For all A,B we have $$\cos(A+B)=\cos A \cos B -\sin A \sin B.$$ $$\sin (A+B)=\sin A\cos B+\cos A\sin B.$$ $$\sin (-A)=-\sin A.$$ $$\cos (-A)=\cos A.$$ $$(\cos A)^2+(\sin A)^2=1.$$ From these, the other "reduction" formulae follow. The trig functions arise naturally in physics where it is often necessary for the domains of the functions to be unlimited.

As a special case of a general principle

Lemma: If $f(t)$ is an analytic function on some interval and is zero on some sub interval $(c,d)$ with $c < d$, then it is zero on the entire interval.

It turns out that addition, subtraction, multiplication, division (when the denominator is nonzero), polynomials, trigonometric functions, logarithms, and exponentiation (where the base is positive) are all analytic functions, and composition of analytic functions gives analytic functions too.

Applying this lemma, the function $\sin(\pi + \theta) + \sin(\theta)$ is an analytic defined for all real $\theta$ (complex too, but that's a diversion). Because it is zero for $0 < \theta < \frac{\pi}{2}$, the lemma tells us that it is zero everywhere.

Thus, we conclude $\sin(\pi + \theta) = -\sin(\theta)$ for all $\theta$.

To use this lemma, we must also understand what isn't analytic. Functions aren't analytic when they're undefined. Square roots (and higher root functions) are not analytic at zero. e.g. when trying to use this lemma to reason with $\sin \theta = \sqrt{1 - (\cos \theta)^2}$, the scope of the lemma stops where the square root is zero; e.g. if we know it's true for $0 < \theta < \frac{\pi}{2}$, then the lemma tells us it's true for all $0 < \theta < \pi$ too, but this lemma doesn't let you go any further than that, because that's where the square root becomes zero.