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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.

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    $\begingroup$ By definition? This really depends on how the functions are defined to begin with. $\endgroup$ – egreg Dec 14 '13 at 22:37
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    $\begingroup$ Do you know pythagorean theorem? $\endgroup$ – user63181 Dec 14 '13 at 22:40
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    $\begingroup$ 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? $\endgroup$ – Nick Dec 14 '13 at 23:10
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    $\begingroup$ 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. $\endgroup$ – Julien Dec 14 '13 at 23:19
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    $\begingroup$ @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. $\endgroup$ – Nick Dec 14 '13 at 23:52

16 Answers 16

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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

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    $\begingroup$ The unit circle definition is just downright beautiful because just by existing it proves the identity. No tricks, no complication, just simplicity. $\endgroup$ – Nick Apr 12 '14 at 16:59
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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 $$f(\theta)=f(0)=1$$

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    $\begingroup$ Sweet.${}{{}{}}$ $\endgroup$ – Git Gud Dec 14 '13 at 22:48
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    $\begingroup$ Bravo! Wow, this is priceless. $\endgroup$ – Nick Dec 14 '13 at 22:53
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    $\begingroup$ 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. $\endgroup$ – Leox Dec 15 '13 at 0:08
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    $\begingroup$ @Leox Series definition. There are other alternatives too. $\endgroup$ – Git Gud Dec 15 '13 at 0:23
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    $\begingroup$ Since $f'$ is equal to zero then $f$ is a constant on $\Bbb R$ so any value gives this constant@AB_ $\endgroup$ – user63181 Feb 23 '15 at 21:02
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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: $$\begin{align} (\cos(\theta))^2+(\sin (\theta))^2&= \dfrac{e^{ 2i\theta}+2+e^{-2i\theta}}{4}-\dfrac{e^{2i\theta}-2+e^{-2i\theta}}{4}\\ &=\dfrac {2-(-2)}4=1.\end{align}$$

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    $\begingroup$ See this is the type of answer I wanted something different and not always thought of (not by highschoolers atleast) Thank you for this. $\endgroup$ – Nick Dec 14 '13 at 22:47
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    $\begingroup$ Glad to have found this! All the other proofs here are based on the Pythagorean theorem or on other trigonometric identities. This one's truly unique! $\endgroup$ – Anindya Mahajan Dec 16 '18 at 0:20
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Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,

triangle ABC

By the Pythagorean theorem, $$ {AC}^2+{BC}^2 = {AB}^2 $$ Dividing by $AB^2$, $$ \require{cancel} \begin{align} &\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\ &\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}}\Big)^2 + \Big(\frac{\text{adjacent}}{\text{hypotenuse}}\Big)^2 = \frac{\cancel{AB^2}}{\cancel{AB^2}} = 1\\ &\Rightarrow \boxed{\sin^2\theta + \cos^2\theta = 1} \end{align} $$

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    $\begingroup$ This is a nice "first proof" to show someone as it only requires the basics of trigonometry; for pedagogical purposes I'd note the disadvantage of this proof is that it only handles acute angles. The unit circle proof is very similar in spirit - indeed you can apply Pythagoras - but does work for all angles. $\endgroup$ – Silverfish Jan 1 '15 at 18:36
  • $\begingroup$ @Silverfish: I agree and it is for that reason, that I had chosen the unit circle answer above all others (even the amazingly good ones). But FYI, the above first proof can be extended for all angles since an obtuse angle can be expressed as the sum of a number of acute angles. It's conceptually simple but geometrically complicated for $\theta > 2\pi$ (but if you apply certain results, you can reduce it back down to simple) and due to its trivial nature, I leave it an exercise for your imagination. $\endgroup$ – Nick Jan 2 '15 at 7:38
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    $\begingroup$ I agree. I'm only noting that pedagogically you probably don't want to use the angle sum formula because that's usually taught later, and doing it diagrammatically gets a bit messy compared to how "clean" the unit circle is. But it's possible. $\endgroup$ – Silverfish Jan 2 '15 at 9:09
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    $\begingroup$ @Silverfish: Hence, we have reached a consensus. All hail $x^2 + y^2 = 1$ $\endgroup$ – Nick Jan 2 '15 at 9:38
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$$\large \sin^2\theta + \cos^2\theta =\sin\theta\sin\theta+\cos\theta\cos\theta =\cos(\theta-\theta) =\cos0 =1$$

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    $\begingroup$ You must know $\sin^2\theta+\cos^2\theta=1$ in order to prove the subtraction formula. $\endgroup$ – egreg Dec 14 '13 at 22:53
  • $\begingroup$ @egreg: I know this is sorta like proving addition using multiplication but this is a proof none the less isn't it? $\endgroup$ – Nick Dec 14 '13 at 23:07
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    $\begingroup$ No, it proves nothing, unless you provide a definition of sine and cosine and show the subtraction formula without using the Pythagorean identity. $\endgroup$ – egreg Dec 14 '13 at 23:24
  • $\begingroup$ @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? $\endgroup$ – Nick Dec 14 '13 at 23:36
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    $\begingroup$ Actually, this is what I was looking for: math.stackexchange.com/questions/3356/… $\endgroup$ – Dylan Yott Dec 14 '13 at 23:50
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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 $1$.

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  • $\begingroup$ This is exactly the solution given by Sami Ben Romdhane above! $\endgroup$ – Idris Dec 19 '14 at 12:54
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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 :)

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    $\begingroup$ 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. $\endgroup$ – Nick Dec 14 '13 at 23:23
  • $\begingroup$ 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. $\endgroup$ – Dylan Yott Dec 14 '13 at 23:31
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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 $$ \begin{align}\cos^2t+\sin^2t&=(\cos t+i\sin t)(\cos t-i\sin t)\\ &=\exp(it)\cdot\overline{\exp(it)}\\ &=\exp(it)\cdot\exp(\overline{it})\\ &=\exp(it)\cdot\exp(-it)\\&=\exp(it-it)\\&=\exp(0)\\&=1.\end{align}$$

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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!

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  • $\begingroup$ 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. $\endgroup$ – R R Dec 14 '13 at 23:57
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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.

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$\mathbb{}$$\mathbb{}$$\mathbb{}$Hint:

enter image description here

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    $\begingroup$ Nice P.S. comment, in the future you can also put \$\mathbb{}$ to fill space when you don't have anything left to say. $\endgroup$ – Joao Oct 22 '14 at 5:51
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On one hand, $$ \int_0^x\sin(x)\cos(x)dx= \int_0^x\sin(x)d(\sin(x))dx= \frac{1}{2}\sin^2(x), $$ On the other hand, $$ \int_0^x\sin(x)\cos(x)dx= -\int_0^x\cos(x)d(\cos(x))dx=- \frac{1}{2}\cos^2(x)+ \frac{1}{2}, $$ Hence, by subtraction, we will have that, $$ 0= \frac{1}{2}\sin^2(x)+ \frac{1}{2}\cos^2(x)- \frac{1}{2} $$ or, equivalently, $$ \sin^2(x)+\cos^2(x)=1. $$ I have not seen this proof elsewhere. It is fun.

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  • $\begingroup$ Welcome. Email and signatures are not to be used here; every post is already signed with your usercard. Regarding your proof, it could be also expressed as $(\sin^2x+\cos^2x)'=0$ combined with $\sin(0)=0$ and $\cos(0)=1$. $\endgroup$ – user147263 Oct 9 '15 at 11:58
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Proof by using Euler's theorem: $e^{i\theta}=(\cos\theta+i\sin\theta)$

We know that $i^2=-1$, hence $$\color{red}{\cos^2\theta+\sin^2\theta}$$ $$=\cos^2\theta-i^2\sin^2\theta$$ $$=(\cos\theta)^2-(i\sin\theta)^2$$ $$=(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)$$ $$=(e^{i\theta})(e^{-i\theta})=e^0=\color{red}{1}$$

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Here are two proofs using only the angle sum identities, the fact the trig functions are periodic, and their values at $0$. It is inspired on the connection to rotations and the fact that rotations don't change the sizes of things, but do not actually assume that connection.

Define the matrix

$$ A(\theta) = \left( \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right) $$

By the angle addition formula, we see that $A(\theta + \varphi) = A(\theta) A(\varphi)$; in particular, $A(n \theta) = A(\theta)^n$.

There are arbitrarily large integer multiples of $\theta$ that are arbitrarily close to integer multiples of $2 \pi$, which gives

$$ A(\theta)^n = A(n \theta) \approx I $$

where $I$ is the identity matrix. By taking determinants, we get

$$ (\cos(\theta)^2 + \sin(\theta)^2)^n \approx 1$$

for arbitrarily large $n$; since the determinant is a real nonnegative number, the only possibility is that

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


To get the details right, we select sequences of integers $a_n, b_n$ with

$$ \lim_{n \to \infty} a_n \theta - 2 \pi b_n = 0 $$ $$ \lim_{n \to \infty} a_n = +\infty $$

e.g. this can be done by continued fractions. Then,

$$ \lim_{n \to \infty} A(a_n \theta) = \lim_{n \to \infty} A(a_n \theta - 2 \pi b_n) = A(0) = I$$

and consequently

$$ \lim_{n \to \infty} (\det A(\theta))^{a_n} = 1 $$


Another method in the same vein is to also define vectors

$$ v(\theta) = \left( \begin{matrix} \sin(\theta) \\ \cos(\theta) \end{matrix} \right) $$

so that $A(\theta) v(\varphi) = v(\theta + \varphi)$

However, we know that for every angle $\theta$:

$$\frac{1}{2} \leq \max(\sin(\theta)^2, \cos(\theta)^2) \leq 1 $$

In particular, we have

$$ \frac{1}{2} \leq \| v(\theta) \| \leq 2 $$

and furthermore, the $v(\theta)$ span $\mathbb{R}^2$.

By comparing the lengths of $v(\varphi)$ and $A(\theta) v(\varphi)$, we know that all of the eigenvalues of $A(\theta)$ must have magnitude lying between $1/2$ and $2$.

However, this remains true for $A(\theta)^n$, and consequently, all of the eigenvalues of $A(\theta)$ must be roots of unity, and thus the determinant is either $1$ or $-1$, and it can't be $-1$.

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Suppose you have a circle with radius r centered at the origin. Using the $ x,y,r$ definitions of $\cos(\theta)$ and $\sin(\theta)$, we have that a point on our circle $(x,y)$, has $x=r\cos(\theta)$ and $y=r\sin(\theta)$. By Pythagorean Theorem, we have $(r\cos\theta)^2+(r\sin\theta)^2 = r^2$ which implies $\cos^2(\theta)+\sin^2(\theta) = 1$

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Here is a nice proof by squaring the infinite Taylor series of $\sin x$ and $\cos x$.

The proof:

$$\sin x=\frac{x}{1}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\\\\\sin^2x=x^2-x^4\left (\frac{1}{1!3!}+\frac{1}{3!1!}\right )+x^6\left (\frac{1}{1!5!}+\frac{1}{3!3!}+\frac{1}{5!1!}\right )-...\\\\\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\\\\\cos^2x\!=\!1\!-\!x^2\left(\!\frac{1}{0!2!}\!+\!\frac{1}{2!0!}\!\right)\!+\!x^4\left(\!\frac{1}{0!4!}\!+\!\frac{1}{2!2!}\!+\!\frac{1}{4!0!}\!\right)\!-\!x^6\left(\!\frac{1}{0!6!}\!+\!\frac{1}{2!4!}\!+\!\frac{1}{4!2!}\!+\!\frac{1}{6!0!}\!\right)\!+...$$We should have shown that the series for both $\sin x$ and $\cos x$ converge absolutely (since we changed the arrangement), but it's obvious since the absolute value of all terms of $\sin x+\cos x$ add up to $e^x$.$$\sin^2x+\cos^2x=\\=1-x^2\left(\frac{1}{0!2!}-\frac{1}{1!1!}+\frac{1}{2!0!}\right)+x^4\left(\frac{1}{0!4!}-\frac{1}{1!3!}+\frac{1}{2!2!}-\frac{1}{3!1!}+\frac{1}{4!0!}\right)-x^6\left(\frac{1}{0!6!}-\frac{1}{1!5!}+\frac{1}{2!4!}-\frac{1}{3!3!}+\frac{1}{4!2!}-\frac{1}{5!1!}+\frac{1}{6!0!}\right)+...=\\\\=1+\sum_{n=1}^{\infty}(-1)^nx^{2n}\sum_{k=0}^{2n}\frac{(-1)^k\binom{2n}{k}}{(2n)!}$$ Since we can show easily that $\sum_{i=0}^n(-1)^i\binom{n}{i}=0$ by expanding $(1-1)^n$ using Binom's formula. So:$$\sin^2x+\cos^2x=1-0+0-0+...=1$$

I think it's beautiful.

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  • $\begingroup$ This is super. I have also seen the associated animation. I like it! $\endgroup$ – Nick Sep 29 '18 at 1:42

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