I want to prove that $\sin(-\theta) = -\sin(\theta)$ and $\cos(-\theta) = \cos(\theta)$ are true without using the summation identities for $\sin(x \pm y)$ or $\cos(x \pm y)$. This is easy enough to do with those formulas if I use $0-x$ as my arguments there.

Without the formulas: Is it usually just by virtue of the unit circle? If I move $-\theta$ around (i.e. clockwise) on the unit circle, I'll end up at some position and the $\sin$ will correspond to that $y$ value. If I had instead moved $\theta$ around counter-clockwise, I'd be at the same point but just the vertical mirror (this is a result of us starting at $(1,0)$ on the $x$-axis), so the $y$ position would be the negative flip of the first one I found. So that one makes sense as to why $\sin(-\theta) = -\sin(\theta)$.

For $\cos$, for the $x$-coordinates, the vertical mirroring doesn't impact this in either case, so $\cos(\theta) = \cos(-\theta)$.

These are a little informal even though it seems to be right, but is there any other way to derive these results without using the summation formula?

  • 1
    $\begingroup$ Then I could be asking the equivalent question "how do we know $\cos$ is even and $\sin$ is odd?" without using the addition formula. $\endgroup$
    – user525966
    Feb 26, 2018 at 4:23
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    $\begingroup$ Typically we use the fact that sine is odd and that cosine is even in order to even obtain the angle addition formulae in the first place. I think that it can be reasonably argued that the parity of these functions is more fundamental than those formulae. That being said, deducing the parity of a trigonometric function depends some on how you define it. If you define it from the unit circle, then that is how you are going to have to obtain the parity. But one could also use definitions that come from power series or differential equations. $\endgroup$
    – Xander Henderson
    Feb 26, 2018 at 4:25
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    $\begingroup$ In order for the question to be meaningful, first we have to define the sine and cosine functions for both positive and negative angles. The unit circle is a good way to do this. With that definition, the result follows for the reasons you gave. $\endgroup$
    – David K
    Feb 26, 2018 at 4:48
  • $\begingroup$ Your argument is correct. Informal means "not formal" and not "non-rigorous". $\endgroup$
    – Paramanand Singh
    Feb 27, 2018 at 4:42

4 Answers 4


Are you familiar with Euler's Formula?

$$e^{i\theta} = \cos(\theta) + i\sin(\theta)$$

By plugging in $-\theta$ to this formula, applying the reciprocal rule of exponents to the left side, simplifying, and then equating the real and imaginary parts will result in what your are trying to prove:

$$e^{-i\theta} = \cos(-\theta) + i\sin(-\theta)$$

$$\dfrac{1}{e^{i\theta}} = \cos(-\theta) + i\sin(-\theta)$$

$$\dfrac{1}{\cos(\theta)+i\sin(\theta)} = \cos(-\theta) + i\sin(-\theta)$$

$$\cos(\theta)-i\sin(\theta) = \cos(-\theta) + i\sin(-\theta)$$

Sorry for the bad quality and limited work, I am doing this on my phone.

  • $\begingroup$ I don't see how you inverted the lefthand side towards the end there $\endgroup$
    – user525966
    Feb 26, 2018 at 5:00
  • $\begingroup$ @user525966 The reciprocal of a complex number is its conjugate over the square of the modulus, or $\cos(\theta) - i\sin(\theta)$ $\endgroup$
    – Andrew Li
    Feb 26, 2018 at 5:02
  • $\begingroup$ From a little more inspection I think you multiplied the lefthand side by $(\cos(\theta) - i \sin(\theta)) / (\cos(\theta) - i \sin(\theta))$ and so the denominator turns to $\cos(\theta)^2 + \sin(\theta)^2 = 1$ $\endgroup$
    – user525966
    Feb 26, 2018 at 5:05
  • $\begingroup$ Nice use of Euler's identity! $\endgroup$
    – Yuriy S
    Feb 26, 2018 at 5:07
  • $\begingroup$ Thanks! One of my calculus teachers taught me early on that most trig identities become really obvious when put in terms of Euler's identity, thus getting rid of the need to memorize then. Plus you usually get 2 identities out of it from the real and imaginary parts. $\endgroup$ Feb 26, 2018 at 12:14

You can also alternatively deduce the parity of the sine and cosine function by their Maclaurin series expansion.

\begin{align} \cos (x)&=\sum_{n=0}^{\infty}(-1)^n \frac{(x^2)^n}{(2n!} \\ \cos(-x)&=\sum_{n=0}^{\infty}(-1)^n \frac{((-x)^2)^n}{(2n!} \\ \cos(-x)&=\sum_{n=0}^{\infty}(-1)^n \frac{(x^2)^n}{(2n!} \\ \implies \cos(x)&=\cos(-x) \\ \\ \sin(x)&=\sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \\ \sin(-x)&=\sum_{n=0}^{\infty} (-1)^{n} \frac{(-x)^{2n+1}}{(2n+1)!} \\ \sin(-x)&=-\sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \\ \implies \sin(-x)&=-\sin(x) \end{align}

  • $\begingroup$ I think for $\sin$ it's $(-1)^n$ still is it not? $\endgroup$
    – user525966
    Feb 26, 2018 at 5:09
  • $\begingroup$ I'm not sure what you mean. $(-1)^{2n+1}=(-1)^{2n}\cdot (-1)^1=-1\quad n\in\mathbb{Z}$. $\endgroup$
    – Diffusion
    Feb 26, 2018 at 5:24
  • $\begingroup$ I mean your equation for $\sin(x)$ is not the usual one that I see because it's using $(-1)^{n+1}$ instead of $(-1)^n$ which I don't believe is right $\endgroup$
    – user525966
    Feb 26, 2018 at 5:47
  • $\begingroup$ Yes, you are right. I edited it. $\endgroup$
    – Diffusion
    Feb 26, 2018 at 6:17

Suppose that we define sine and cosine using the unit circle as follows:

Definition: An angle $\theta \in [-\pi, pi)$ in standard position, i.e. measured anti-clockwise from the point $(1,0)$ on the unit circle, intersects the unit circle at a unique point $(x_{\theta},y_{\theta})$. We define $$ \cos(\theta) := x_{\theta} \qquad\text{and}\qquad \sin(\theta) := y_{\theta}. $$

This is slightly informal, but is generally the way things are done in your typical precalculus class. Note that I have restricted my angles to measures between $-\pi$ and $\pi$ (picking one of the two endpoints more-or-less arbitrarily). We could count the number of times that we "wrap around" the circle in order to measure larger or smaller angles, but the basic facts won't actually change if we do this, so I'll stick with the simplification that I've made and leave it as an exercise to show that the "extended" sine and cosine functions retain their parities.

Since the definition is geometrical in nature, let's use some geometry to answer the question. Fix some angle $\theta \in [0,\pi)$ and define \begin{align} A &= (x_{\theta}, y_{\theta}) = (\cos(\theta), \sin(\theta)), \\ A' &= (x_{-\theta}, y_{-\theta}) = (\cos(-\theta), \sin(-\theta)), \\ B &= (1,0), \qquad\text{and}\\ O &= (0,0). \end{align} The basic construction can been seen and played with a bit in this Desmos demonstration (move the slider to change the angle). Honestly, the demonstration is pretty convincing on its own. Everything from here on out is basically a pedantic, semi-rigorous argument. Feel free to stop now if you are convinced.

Consider the triangles $\triangle OAB$ and $\triangle OA'B$. By construction, $\angle AOB \cong \angle A'OB$, since both angles have measure $\theta$. Also, the triangles share a side (namely, $\overline{OB}$) and $\overline{OA} \cong \overline{OA'}$, as each is a radius of the unit circle. Via the side-angle-side congruence relation, this implies that the two triangles are congruent.

Now consider dropping perpendicular segments from $A$ and $A$, resp., to the $x$-axis. These segments are the heights of $\triangle OAB$ and $\triangle OA'B$, resp., and must be congruent as the two triangles are congruent. But this means that the $x$-coordinate of the point where the heights intersect the $x$-axis must be the same, namely the point $(x_\theta, 0) = (x_{-\theta},0)$. In other words, we can use this construction to show that $\cos(\theta) = \cos(-\theta)$ for all $\theta \in [0,\pi)$ (and use symmetry to get the rest of the circle). In any event, the cosine function is therefore even.

Also, since the heights are congruent, but oriented on opposite sides of the $x$-axis, we see that $y_{-\theta} = -y_{\theta}$, i.e. $\sin(-\theta) = -\sin(\theta)$. After a small argument, this implies that the sine function is odd. Therefore we may obtain the parities directly from the definitions, assuming that we define the functions geometrically.


As I noted in my comment to the original question, deducing basic properties of trigonometric functions depends deeply on how you define them. In my other answer, I attacked the problem as might be done in a precalculus class. Messney provided a solution via power series expansion (one must assume that they are taking the power series as the definition), and Aaron Stevens used complex analysis (again, we must assume that he is taking the Euler formula as the definition; though how are the exponentials defined?). Here is another alternative approach, using differential equations.

Proposition 1: The initial value problem $$ \begin{cases} u'' + u = 0 \\ u(0) = a \\ u'(0) = b \end{cases} \tag{$\ast$} $$ has a unique solution for any choice of $a,b\in\mathbb{R}$.

I am going to take this proposition on faith. As is often the case, the uniqueness result is not horrible to obtain, but the existence result takes some work. If you are willing to take this on faith, then define $\sin$ to be the solution of ($\ast$) corresponding to the initial condition $$ \sin(0) = 0 \qquad\text{and}\qquad \sin'(0) = 1, $$ and define $\cos$ to be the solution of ($\ast$) corresponding to the initial condition $$ \cos(0) = 1 \qquad\text{and}\qquad \cos'(0) = 0. $$ I should probably call these functions "$s$" and "$c$", because it is not immediately obvious that they are the same as the usual sine and cosine functions. Indeed, the three propositions below are only the first steps in establishing this fact. However, if you've gotten this far, you are already taking quite a bit on faith, so what's a bit more?

We now get the first basic property of sine and cosine, namely that they are related to each other via differentiation in the usual manner.

Proposition 2: The following identities hold: $$ \sin' = \cos \qquad\text{and}\qquad \cos' = -\sin. $$

Proof: By definition $\sin$ solves ($\ast$). Differentiating, we obtain $$ (\sin'' + \sin)' = \sin''' + \sin' = (\sin')'' + (\sin') = 0. $$ Therefore $\sin'$ is a solution to ($\ast$). Moreover, by definition $\sin'(0) = 1$, and it follows from ($\ast$) that $$ \sin'' + \sin = 0 \implies \sin''(0) = -\sin(0) = 0. $$ Therefore $\sin'$ is a solution to the IVP ($\ast$) satisfying the initial conditions $\sin'(0) = 1$ and $(\sin')'(0) = 0$. But then, by definition of the cosine function, we have $$ \sin' = \cos. $$ By a similar argument, we may show that $\cos' = -\sin$.$\tag*{$\square$}$

This now gives us enough to show that the cosine function is even.

Proposition 3: The function $\cos$ is even; that is, we have $\cos(-t) = \cos(t)$.

Proof: Define $u(t) = \cos(-t)$. Then observe that \begin{align} u''(t) &= \frac{\mathrm{d}^2}{\mathrm{d}t^2} \cos(-t) \\ &= \frac{\mathrm{d}}{\mathrm{d}t}\sin(-t) \tag{Prop. 2 and chain rule} \\ &= -\cos(-t) \tag{Prop. 2 and chain rule} \\ &= -u(t). \end{align} Moreover $u(0) = \cos(0) = 1$ and $u'(0) = \sin(0) = 0$. But then $u$ solves the IVP ($\ast$) with the same initial conditions as $\cos$. Therefore by the uniqueness of solutions to ($\ast$), we have $$ u(t) = \cos(-t) = \cos(t). $$ Therefore $\cos$ is even.$\tag*{$\square$}$

Proposition 4: The function $\sin$ is odd; that is, we have $\sin(-t) = -\sin(t)$.

Proof: This follows immediately from Proposition 3: since $\cos(-t) = \cos(t)$, we can differentiate (remembering to apply the chain rule and Proposition 2) in order to obtain $$ \cos(-t) = \cos(t) \implies \frac{\mathrm{d}}{\mathrm{d}t} \cos(-t) = \frac{\mathrm{d}}{\mathrm{d}t} \cos(t) \implies \sin(-t) = -\sin(t), $$ which is the desired result.$\tag*{$\square$}$


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