# Integer multiples of $2\pi$ on $\cos$ function

Suppose you know that the smallest positive value $t$ such that $\cos t=0$ is $t=\dfrac{\pi}{2}$, but you don't know other values of $\cos$ and $\sin$.

You also know that $u$ is a real number such that $\cos u=1$ and $\sin u=0$.

Just from this, can you conclude that $u$ must be an integer multiple of $2\pi$?

(You know the usual trig identities, such as $\cos^2x+\sin^2x=1$ and $\cos 2x=2\cos^2x-1$.)

• Ok, what do we know about $\cos$ and $\sin$? Commented Sep 30, 2013 at 0:03
• @Berci Any trig identity you want, such as the ones I included. Commented Sep 30, 2013 at 0:09

Suppose that $\sin \alpha = \sin \beta = 0$. Then $$\sin(\alpha+\beta)=\sin \alpha \cos \beta + \sin \beta \cos \alpha=0,$$ which implies that the zeroes of sine are closed under addition. Since sine is an odd function, they are also closed under additive inverses. It follows that the zeroes of sine form an abelian group.

If this group is not cyclic (generated by a single element), then it is dense in the real line. But, since sine is continuous, it would follow that sine is identically zero, which contradicts that cosine has a zero at $\pi/2$ (using the Pythagorean identity and your hypothesis). Therefore the zeroes of sine are cyclic, which implies that they have a unique minimal positive element. It remains to show that this element is $\pi$. Using the double angle formula for cosine, $\cos 2 x = 2\cos^2 x -1,$ it follows that $\cos \pi =-1$, so that $\sin \pi=0$. On the other hand, suppose that sine has a minimal root $x \in (0,\pi)$. Then $\cos x = \pm 1$, so that $$\cos x/2 = \pm \sqrt{\frac{1+\cos x}{2}}=0 \text{ or } \pm1.$$ (This is, again, the double angle law for cosine.) Then $\cos x/2 =1$ since cosine is continuous, $\cos 0=1$, and $x/2 < \pi/2$, the minimal positive zero of cosine. It follows that $\sin(x/2)=0$, which contradicts that $x$ was chosen minimally. So $x \geq \pi$, and $x = \pi$ as seen above.

This answers half of your question. It remains to study the behavior of cosine on integer multiples of $\pi$. Suppose that $\cos \alpha=\cos \beta =1$. Then $\alpha,\beta$ are zeroes of sine, and the angle addition law for cosine gives $$\cos (\alpha+\beta)= \cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos \alpha \cos \beta =1.$$ As well, since cosine is an even function, it follows that the inverse image $$G=\{x \in \mathbb{R} : \cos x =1\}$$ is closed under additive inverses, hence is a subgroup of $\pi \mathbb{Z}$. This subgroup does not contain $\pi$, since $\cos \pi =2 \cos^2(\pi/2)-1=-1$. $G$ does contain $2\pi$, however, since $\cos 2\pi = 2 \cos^2\pi-1=1$. Thus $G$ is generated by $2\pi$, and the set of $x$ such that $\sin x =0$ and $\cos x=1$ is precisely $2\pi \mathbb{Z}$.

• This may be a bit much for a question tagged (algebra-precalculus). I would assume the "all we know are the trig identities" precludes us from using group theory and such. Commented Sep 30, 2013 at 1:04
• In counterpoint, I might argue that the soft induction in your proof rises above the level of pre-calculus. I will leave my answer as it stands, in case others wish to see it. Commented Sep 30, 2013 at 1:34

All we know is the trig identities and that $\cos\theta>0$ when $0<\theta< \frac{\pi}{2}$ and $\cos(\frac{\pi}{2})=0$.

Now, $\cos( \frac{\pi}{2} - \theta) = \sin \theta$ tells us $\sin \theta>0$ when $0 < \theta < \frac{\pi}{2}$ and $\sin(0)=0$.

From $\sin(2 \theta) = 2\cos \theta \sin \theta$ we see $\sin \theta >0$ for $0<\theta<\pi$ and $\sin(\pi)=0$.

Continuing like this, and then using $\sin(-\theta)=-\sin\theta$, you will be able to generate a complete list of all the zeroes of $\sin$. Spoiler: the integer multiples of $\pi$. Now you're halfway there, but you also need a list of where $\cos\theta=1$.

From $\cos(2\theta)= 1 - 2\sin^2(\theta)$ you see that $\cos\theta=1$ at precisely the doubles of the zeros of $\sin$. But you've found that the zeroes of $\sin$ are the integer multiples of $\pi$. So you're done. Finally!

Note that we'll never need to assume $\sin$ or $\cos$ are continuous, which is appropriate for a question tagged "precalculus."