Ok, so I know that $\sin^2(x)+\cos^2(x)=1$ for all angles. If $x$ is an acute angle in a right angled triangle it's a straightforward proof. But what about if $x$ is obtuse ? How do I mathematically prove it plus get a visual analysis of the same so you can use it in $2D$ or $3D$ geometry using the co-ordinate axes.

Edit: I know the proofs in which they show a unit circle. But I thought the obtuse angles were only designed to incorporate the sign. For example say a point on the unit circle is (-0.5,0.866). This shows up for x=120. However, we do the calculations for the acute angle from the negative x axis and then just put sign for sin(x) or cos(x) if x is any angle so that when we resubstitue x=r*sin(x) we get the polarity of x because sin(x) can be positive or negative but r is always considered postive. r is the distance from the origin.

But what if we have a triangle in space which has coordinates say (1,1),(1,5) and (-2,8). How do we use trigonometry in those cases ? Since the angle made with the x axis for the above points doesn't matter since the angle between the linea joining those points are something totally different

  • 1
    $\begingroup$ Hint: Think about how to interpret this situation on the unit circle. $\endgroup$ – David Kraemer Sep 27 '18 at 14:32
  • 2
    $\begingroup$ What are your definitions of sine and cosine? $\endgroup$ – Mark Sep 27 '18 at 14:32
  • $\begingroup$ If you used only pythagorean theorem, you can use $\sin(x+y)=\sin x\cos y+\sin y\cos x$. Thus, for a obtuse angle use this with $\theta=\pi/2+\alpha$, where $\alpha$ is a acute angle. Then, use again pythagorean theorem. $\endgroup$ – DiegoMath Sep 27 '18 at 14:34
  • $\begingroup$ In order to prove this for obtuse angles you first need a definition for sine and cosine of obtuse angles. Do you have one? $\endgroup$ – Arthur Sep 27 '18 at 14:34
  • 1
    $\begingroup$ @NigelGoveas That definition is not going to help for obtuse angles. You need a definition which works for obtuse angles if you want to prove anything for obtuse angles. $\endgroup$ – Arthur Sep 27 '18 at 15:42

Assume $\pi/2 < \theta < \pi$. Then $\sin(\theta) = \sin(\pi-\theta)$ and $\cos(\theta) = -\cos(\pi-\theta)$ and so you can reduce to the case of the acute angle $\pi-\theta$.

  • $\begingroup$ The question is does OP know how to prove the two identities you wrote. $\endgroup$ – Mark Sep 27 '18 at 14:36

A complex analysis reasoning:

Let $$f(z)=\sin^2(z)+\cos^2(z)$$ $$g(z)=1$$

Since $f=g$ for all $z\in[0,\frac{\pi}2]$ and $f,g$ are holomorphic on $\mathbb C$, $f=g$ for all $z\in\mathbb C$ by identity theorem.


Here are two answers, each based on a different definition.

Answer 1: We define, for acute angles with one ray on the positive $x$-axis and the other in the first quadrant, the sine and cosine as the $y$ and $x$ coordinates of the point where the second ray meets the unit circle. This gives us a notion of sine and cosine of $x$ for $0 \le x \le \pi/2$.

For obtuse angles $\pi/2 < x \le \pi$, we define $\sin(x) = \sin(\pi - x)$ and $\cos(x) = -\cos(\pi - x)$. We verify (this takes a little work) that sine and cosine, even for these new angles, satisfy $\sin^2(x) + \cos^2(x) = 1$, and that they're continuous at $x = \pi/2$, and that the two definitions agree there.

[Let me do that work: suppose that $\pi/2 < x /le \pi$; then $\sin(x) = \sin(\pi-x)$, where the latter argument is a number $y = \pi - x$ between $0$ and $\pi/2$; similarly, $\cos(x) = -cos(y)$. So $\sin^2(x) + \cos^2(x) = [\sin(y)]^2 + [-\cos(y)]^2 = \sin^2(y) + \cos^2(y)$, where $y$ is an acute angle, so this sum is $1$.]

We then do something similar for angles between $\pi$ and $3\pi/2$ (defining $\sin(x) = -\sin(x-\pi), \cos(x) = -\cos(x-\pi)$), and for angles between $3 \pi/2$ and $2\pi$ (I'll let you fill in the definition yourself).

Then we define sine and cosine for an arbitrary angle $a$ in any triangle $ABC$ by observing that there's a unique translation and rotation that moves the ray $AB$ to the positive $x$-axis, and the ray $AC$ to ...some ray starting at the origin. You can then use the earlier definition of sine and cosine to define the sine and cosine of $a$. And that finishes things up.

Of course, proving the existence an uniqueness of a translation $T$ and rotation $R$ with the property that $R(T(A)) $ is the origin, etc., involves some real work, which is why I hate this approach.

Approach 2: There are, on the rationals, two unique functions, $s$ and $c$, with the following four properties:

  1. $s(0) = 0; c(0) = 1; s(\pi/2) = 1.$

  2. For $0 < x < \pi/2$, $0 < s(x) < x$.

  3. $c(a-b) = c(a) c(b) + s(a) s(b)$ for all rational $a, b$.

  4. $s$ and $c$ are continuous on the rationals.

[I think I have those right, but I might be mistaken; I'm pretty sure they're in Apostol's Calculus book]

Proving the existence and uniqueness of these two functions requires some real work. You basically use formula 3 to derive a cosine double-angle formula, and then show that the facts in item 1 (plus formula 3) let you compute sines and cosines of things like $\pi/2, \pi/4, 3\pi/4, \pi/8, ...$, basically, $\pi$ times any rational whose denominator is a power of two. Then you use a limit argument (this relies heavily on item 2) to conclude that the value on all rationals must be determined because of property 4.

THEN, you say "every continuous function on a dense subset of the reals extends to a unique continuous function on the reals" (which requires proof, and a notion of "dense"!), and you say that we'll call the unique continuous extension of $s$ by the name "sine", and the extension for $c$ by the name "cosine". Finally, plugging in $a = b = x$ for any rational $x$ in formula $3$, you see that $$ \cos(0) = \cos(x)^2 + \sin(x)^2, $$ and since, by rule 1, we know that $\cos(0) = 1$, we get that the fundamental identity holds on the rationals, which is to say, the function $$ u(x) = \cos^2(x) + \sin^2(x) - 1 $$ is zero for every rational $x$. It's also continuous, so it has a unique continuous extension to the reals, namely, the everwhere zero function. So for every REAL number $x$, we have $$ 0 = \cos^2(x) + \sin^2(x) - 1 $$ i.e., the fundamental identity is proved.

Of course, although the mathematics here is not terribly difficult, it's really not immediately clear where the four "axioms" came from, or why the functions $s$ and $c$ have anything to do with the things we're used to calling "sine" and "cosine".

There's one more version of this that's possible, which involves defining sines and cosines using matrices (or complex numbers), but I'm not sure that approach is likely to be much more enlightening than these.

The short summary is "This is actually somewhat challenging to do rigorously, which is why basic courses tend to gloss over it."

  • $\begingroup$ Now this was a great answer. I'm going to read Apostol's Calculus book right away $\endgroup$ – Nigel Goveas Sep 28 '18 at 22:26
  • $\begingroup$ Wow...well, I sure hope that this IS in Apostol. I learned it all from a mimeographed geometry text by King and Burdoin that was used at a prep school back in the 1970s, but I'm pretty sure that they didn't invent it. $\endgroup$ – John Hughes Sep 29 '18 at 0:52

I can't think of a geometric interpretation right now, but I would argue with the angle reduction formulas.

Let $x \in [\frac{\pi}{2}; \pi]$ be an obtuse angle. This means $x$ can be rewritten in terms of an acute angle $y \in [0; \frac{\pi}{2}]$ plus $\frac{\pi}{2}$, meaning $x=y+\frac{\pi}{2}$. Then, with the reduction formulas it holds that $\sin(x)=\sin(y+\frac{\pi}{2})=\cos(y)$ and $\cos(x)=\cos(y+\frac{\pi}{2})=-\sin(y)$.

Plugging this in yields


The last equality being true, because of $y$ being an acute angle.

  • $\begingroup$ While writing this, I didn't see that another user already reasoned this way. My answer is just more elaborate, but basically states the same. $\endgroup$ – Korosensei Sep 27 '18 at 14:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.