The question is the following: what are the principal reasons to define the sine, cosine, etc., of an angle, in terms of the right triangle and not, for example, in terms of an obtuse triangle?

1) Is the pythagorean theorem a good reason for using a right triangle? Why? Also, is it true that the ratios of sides of a triangle are always functions of the angle, or this is only true in right-angled ones?

2) Also, I think that, if we define the trigonometric functions in terms of ratios of the sides, we need the right triangle, because it's the only triangle for which their sides can be localized. For example, if we have an obtuse or acute triangle, we cannot determine what is the adjacent side to the angle $A$, because there are two sides who satisfies that property, but in the right-angled triangle, the hypotenuse is always bigger and hence clearly distinguishable of the adjacent side of the angle $A$. Is this a good reason to prefer the right angles triangles to define the trigonometric ratios?

Also, every triangle can be decomposed into two right angled triangles. Is this also a good reason to prefer right triangles, or also any triangle can be decomposed into an obtuse triangle?


2 Answers 2


Given an acute angle $\alpha$, there are infinitely many acute and obtuse triangles having $\alpha$ as one of their angles, even up to similitude. To the contrary, there is only one right triangle, up to similitude, having $\alpha$ as one of its angles.

This caters for the choice when the angle $\alpha$ is acute: the right triangle is the obvious one.

How can we do for obtuse angles? We try and generalize.

If we look at the sine law for acute angles, it turns out easily that we can extend it also to obtuse triangles by defining $$ \sin\beta=\sin(\pi-\beta) $$ if $\beta$ is obtuse.

For the cosine, we can use the cosine law. Let $ABC$ be a triangle with the angle $\beta$ in $B$ obtuse. Draw the height from $A$ relative to $BC$, calling $H$ its foot (which is external to the triangle). Now we have two right triangles: $AHC$ and $AHB$. If $AB=c$, $AC=b$, $BC=a$, $AH=h$ and $HB=d$, we can say $$ c^2=h^2+d^2,\qquad b^2=h^2+(d+a)^2 $$ so, eliminating $h$, $$ b^2=c^2-d^2+(d+a)^2=a^2+c^2+2ad $$ By the definition of cosine, $d=c\cos(\pi-\beta)$, so $$ b^2=a^2+c^2+2ac\cos(\pi-\beta) $$ and it is natural to define $\cos\beta=-\cos(\pi-\beta)$, so the relation becomes $$ b^2=a^2+c^2-2ac\cos\beta $$ which is exactly the same as the cosine law for acute triangles.

The law of cosines applied formally when the angle $\beta$ is right forces us to define $\cos(\pi/2)=0$. The law of sines gives instead $\sin(\pi/2)=1$.

Nowadays, sine and cosine are better defined with the trigonometric circle, which relieves from such geometric considerations which only allow to define sine and cosine for angles in the range $[0,\pi]$: with the trigonometric circle it's easy to define the sine and cosine for any angle, even outside $[0,2\pi]$.

It's also possible to define sine and cosine without any resort to geometry, with their Taylor series.

  • $\begingroup$ Nice answer. It is not my intention to discuss the best way to define trigonometric functions (as, for example, using the unit circle), but how the old mathematicians, who did not have the analytical apparatus of the coordinate system, decided on the right triangle . What I am asking is: is the right triangle the only triangle in which these functions can be defined geometrically, in the sense that it is chosen not only for its advantages, but for being the only option? That is,is it impossible to define the trigonometric ratios in an obtuse or acute triangle for the reason I said? $\endgroup$
    – user397987
    Dec 12, 2016 at 1:14
  • $\begingroup$ @user231312 The problem is ”what obtuse triangle do you choose?“ And how do you prove the definition is invariant? I'm not saying it's impossible, but would you try and check the invariance if you define the cosine with the ”inverse“ cosine law? $\endgroup$
    – egreg
    Dec 12, 2016 at 8:58

I think that the principal reason for that definition is due to the fact that sine and cosine of an angle are, in general, defined by means the unit circle, i.e. the circle of radius one centered in the origin $O(0,0)$.

If we take the unit circle, a radius makes an angle $\alpha$ respect to the positive $x-axis$ and it intersects the circle in a point $P$ with coordinates $(x_P,y_P)$. We can define the sine and cosine of the angle $\alpha$ as the coordinates of the point $P$: $x_P\equiv \cos\alpha$, $y_P\equiv\sin\alpha$. It is obvious that the triangle $\triangle OPH$ is a right triangle (where $H$ is the intersection point between the height $PH$ and the $x-axis$).

So you can use this definition in any right triangle $\triangle OP'H'$ similar to $\triangle OPH$, hence for the cosine we have: $OH:OP=OH':OP'\longrightarrow x_P:1=x_{P'}:OP'\longrightarrow \cos\alpha=x_P={x_{P'}\over OP'}$, that is the "geometric definition" of cosine; similarly for the sine.

From this definition we get the "famous" trigonometric identity just applying the Pythagoras's theorem to the triangle $\triangle OPH$:

$$x_P^2+y_P^2=1\longrightarrow \cos^2\alpha+\sin^2\alpha=1$$


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