Why does $\sin(t) + \cos(t)$ itself look like a sine graph?

Question

So the other night I was randomly python scripting. I plotted $\sin(t) + \cos(t)$ vs $t$ for $t$ ranging between $0$ to $100$ with spacings of Δt = 0.1. (It is a pretty basic code...) Anyhow, the below plot is the result.

Why does it look like this? I cannot figure out why, surely it has something to do with some periodicity going on in $\sin(t) + \cos(t)$?

Hint: $\sin t+\cos t=\sqrt2\left(\frac1{\sqrt2}\sin t+\frac1{\sqrt2}\cos t\right)$. Express $1/\sqrt2$ in trig form. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Aug 23 '21 at 08:42
Got it! In the end it is a sine function with the trig form substitution. — innating, Aug 23 '21 at 08:51
Both the identities for sine and cosine work. You can check that both forms are the same because $\sin(x) = \cos(\pi/2 - x)$, where this is true from the unit circle. — Toby Mak, Aug 23 '21 at 08:55

score 6 · Accepted Answer · answered Aug 23 '21 at 08:46

Recall that by sum/difference identities

$$\cos (\alpha-\beta) =\cos \alpha\cos \beta + \sin \alpha\sin \beta$$

and by $\cos \beta=\sin \beta=\frac{\sqrt 2}2$ we have

$$\cos \alpha + \sin \alpha=\sqrt 2 \cos (\alpha-\beta)$$

with $\beta=\frac{\pi}4+2k\pi$.

score 3 · Answer 2 · answered Aug 23 '21 at 08:53

3

The angle addition formulas are useful, but you don't need them to show that $f(t) = \sin t + \cos t$ is periodic. Since sin and cos are both periodic with a period of $2 \pi$, their sum must be periodic as well, or more formally:

$$f(t - 2 \pi) = \sin(t - 2\pi) + \cos(t - 2 \pi) = \sin t + \cos t = f(t)$$

answered Aug 23 '21 at 08:53

Toby Mak

16,827

Interesting answer. I chose the other answer as the correct one (sorry!) but on another day I might have chose this one. – innating Aug 23 '21 at 23:42

Why does $\sin(t) + \cos(t)$ itself look like a sine graph?

2 Answers2