Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do you prove the following trigonometric identity: $$ \sin^2\theta+\cos^2\theta=1$$

I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.

share|improve this question
4  
By definition? This really depends on how the functions are defined to begin with. –  egreg Dec 14 '13 at 22:37
2  
Do you know pythagorean theorem? –  Sami Ben Romdhane Dec 14 '13 at 22:40
6  
Please don't close this question. Tell me what I should add. Where I can improve my question. Why is there such a big fuss over this? –  Nick Dec 14 '13 at 23:10
3  
Close voters: according to the edit "I'm curious to know of the different ways mathematicians approach this kind of question", I highly doubt this is no effort homework. –  1015 Dec 14 '13 at 23:19
2  
@anorton: You will not believe how many fundamental things are not asked to be proved on M.SE ... But now I'm afraid if I ask those things they will be treated the same way as this question was. –  Nick Dec 14 '13 at 23:52

11 Answers 11

up vote 7 down vote accepted

Let $(\mathscr C)$ be a unit circle, and $\mathrm M\in(\mathscr C)$. Also, we will denote $\rm \angle{IOM}$ as $\theta$ (see the diagram). From the unit circle definition, the coordinates of the point $\rm M$ are $(\cos\theta,\sin\theta)$. And so, $\rm \overline{OC}$ is $\cos \theta$ and $\rm \overline{OS}$ is $\sin \theta$. Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$

$\phantom{X}$unit circle

share|improve this answer
1  
The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. –  Nick Apr 12 at 16:59

Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$

share|improve this answer
4  
Sweet.${}{{}{}}$ –  Git Gud Dec 14 '13 at 22:48
1  
Bravo! Wow, this is priceless. –  Nick Dec 14 '13 at 22:53
2  
I think it is a wrong solution. To prove the formulas $(\sin(x))'=\cos(x)$ and $(\cos(x))'=\sin(x)$ we have to know the main trigonometric identity. –  Leox Dec 15 '13 at 0:08
6  
@Leox Series definition. There are other alternatives too. –  Git Gud Dec 15 '13 at 0:23
2  
Excellent! ${}{}{}{}+ 1$ –  amWhy Dec 15 '13 at 12:56

Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions:

$$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$

Take $\theta \in\mathbb R$. The following holds: $$\begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}$$

share|improve this answer
    
See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. –  Nick Dec 14 '13 at 22:47

Consider a right-angled triangle ABC where $<BAC = \theta$,

triangle ABC

By Pythagorean theorem, $$\Large AC^2+BC^2=AB^2$$ Dividing by $AB^2$, $$\Rightarrow \large\frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}$$ $$\Rightarrow \large(\frac{\text{opposite}}{\text{hypotenuse}})^2+(\frac{\text{adjacent}}{\text{hypotenuse}})^2 = \frac{AB^2}{AB^2}$$ $$\Large\Rightarrow \sin^2\theta+\cos^2\theta=1$$

share|improve this answer

In the spirit of Git Gud's answer, differentiate $\sin^2 \theta + \cos^2 \theta$ to get

$$ 2 \sin \theta \cos \theta - 2 \cos \theta \sin \theta = 0$$

So $\sin^2 \theta + \cos^2 \theta$ is constant. Plugging in $\theta = 0$ shows that constant is $1$.

share|improve this answer

Well it comes directly from the Pythagorean theorem. We know that in a right triangle, $cos {\theta}=\frac{h}{r}$ and $sin{\theta}=\frac{v}{r}$, $h$ is short for horizontal and $v$ for vertical, $r$ is the hypotenuse.

Now, from the Pyth. theorem

$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$

By the way, the Pythagorean theorem is one of the oldest theorems of mathematics. Archaelogists have discovered it inscribred in stones in excavations in Babylon!

share|improve this answer
    
This identity is true for values of $\theta$ that are both smaller than zero and larger than 180 degrees which are not usually seen inside a right triangle. I feel there should some extra justification added when appealing to Pythagoras. –  R R Dec 14 '13 at 23:57

If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few:

$(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative.

Proof of identity using $(1)$: $(\sin^2(x)+\cos^2(x))'=(y^2+y'^2)'= 2yy' + 2y'y''= 2yy'-2yy'=0$, now letting $x=0$ gives the identity. This is similar to Isaac's answer.

$(2)$, $\sin(x)= x-\frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ and $\cos(x)$ is its derivative.

Proof of identity using $(2)$: Define $e^{x}$ by its power series. Now show $e^{ix}=\cos(x)+i\sin(x)$, and use Git Gud's answer.

As you can see, these proofs are related, so its all a matter of definitions. I hope that helps :)

share|improve this answer
1  
hugs At first I thought you weren't very nice because you were shoving me into a corner but now I think you're absolutely splendid for teaching me something I didn't know. Thank you. –  Nick Dec 14 '13 at 23:23
    
Glad I could help! Sorry for seeming a bit mean at first, that was completely unintentional. I do like this question and think its important to see that the different proofs of this fact come from the fact that there are different definitions of sine and cosine. –  Dylan Yott Dec 14 '13 at 23:31

$$\Large\sin^2\theta + cos^2\theta =\sin\theta\sin\theta+cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$

share|improve this answer
8  
You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. –  egreg Dec 14 '13 at 22:53
    
@egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? –  Nick Dec 14 '13 at 23:07
    
No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. –  egreg Dec 14 '13 at 23:24
    
@egreg: Ah yes, I've been meaning to ask someone that. Can I say in my question that it's ok to assume either definition of sine and cosine in answering? –  Nick Dec 14 '13 at 23:36
2  
Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… –  Dylan Yott Dec 14 '13 at 23:50

Going from the opposite/hypotenuse and adjacent/hypotenuse definitions:

Let $\theta\in\left[0,\frac{\pi}{2}\right]$ be an angle (in radians, of course) in a right triangle. Let $a$ be the length of the side of a triangle opposite from the angle $\theta$, $b$ the length of the side adjacent to the angle, and $c$ the length of the hypotenuse. Then, $$\sin^{2}\theta+\cos^{2}\theta=\left(\frac{a}{c}\right)^{2} + \left(\frac{b}{c}\right)^{2} = \frac{a^{2}}{c^{2}}+\frac{b^{2}}{c^{2}}=\frac{a^{2}+b^{2}}{c^{2}}=\frac{c^{2}}{c^{2}}=1.$$

To get this result for $0\leq\theta\leq 2\pi$, note that the higher angles only determine the sign of $\sin$ and $\cos$ when a right triangle is formed by going out some length $c$ at angle $\theta$ in the plane and dropping a line perpendicular to the $x$-axis, and since the sign of $\sin$ and $\cos$ don't matter when squaring, the result still holds. To extend the result further to all $\theta\in\mathbb{R}$, note that we just extend the values of $\sin$ and $\cos$ with period $2\pi$ so that we can use any $\theta\in\mathbb{R}$, and it holds trivially.

share|improve this answer

We can define(!) the (first only $\mathbb R\to\mathbb R$) functions $\sin$ and $\cos$ via $\exp(it)=\cos t+i\sin t$ and the (complex) exponential as unique(!) solution of the differential equation $f'(z)=f(z)$ with $f(0)=1$. We need only a few properties of $\exp$ that quickly follow from uniqueness of the solution:

  • Since $z\mapsto\frac1{\exp a}\exp(z+a)$ is also a solution whenever $\exp(a)\ne 0$, we conclude by uniqueness that $\exp(a+b)=\exp(a)\exp(b)$ whenever $\exp(a)\ne0$.
  • Specifically, $\exp(a)=0$ implies $\exp(a/2)=0$, hence $\exp(2^{-n}a)=0$. As $\exp(0)\ne0$ and $2^{-n}a\to 0$ and $\exp$ is continous, we conclude $\exp(a)\ne0$ for all $a$. Therefore $\exp(a+b)=\exp(a)\exp(b)$ for all $a,b$.
  • Since $z\mapsto\overline{\exp(\overline z)}$ is also a solution, we conclude $\exp\overline z =\overline{\exp z}$ for all $z$.

This makes $$ \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}$$

share|improve this answer

Hint:

enter image description here

PS. Why body must be at least 30 characters ??? It's pointless here.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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