# In how many different ways can you prove that $\sin^2x + \cos^2x = 1$

The standard proof of the identity $$\sin^2x + \cos^2x = 1$$ (the one that is taught in schools) is as follows: from pythagoras theorem, we have (where $$h$$ is hypotenuse, $$b$$ is base and $$p$$ is perpendicular) $$h^2 = p^2 + b^2$$ dividing by $$h^2$$ on both sides: $$1 = \frac{p^2}{h^2}+\frac{b^2}{h^2}$$ since $$\sin x = \frac ph$$ and $$\cos x = \frac bh$$, $$1 = \sin^2x+\cos^2x$$

Are there any more innovative ways of proving this common identity?

• It all boils down to the definition of sin and cos. Which one are you using? – lhf Apr 26 at 14:38
• I bet there are dozens, depending on what you allow to be a given fact. – Marius S.L. Apr 26 at 14:38
• I would guess $n+1$. – Michael Hoppe Apr 26 at 14:52
• – J. W. Tanner Apr 26 at 15:06

Here's a couple of different methods:

Proof using the cosine angel sum formula:$$1=\cos(0) =\cos(x + (-x)) =\cos(x)\cos(-x) - \sin(x)\sin(-x) = \cos^2(x) + \sin^2(x)$$ Proof using differential equations: Recall the differential equation definitions of $$\sin$$ and $$\cos$$: They are the solutions to $$f'' = - f$$ with the appropriate initial conditions $$f(0) = 0$$, $$f'(0) = 1$$ for $$\sin(x)$$ and $$f(0) = 1$$, $$f'(0) = 0$$ for $$\cos(x)$$. Since solutions to differential equations are unique given the initial conditions, we immediately get $$\sin'(x) = \cos(x)$$ under this definition. Then:$$\begin{eqnarray} \frac{d}{dx} \left(\sin^2(x) + \cos^2(x)\right) &=& \frac{d}{dx} \left(\sin^2(x) + \left(\sin'(x)\right)^2\right)\\ &=& 2\sin(x)\sin'(x) + 2\sin'(x)\sin''(x)\\ &=& 2\sin(x)\sin'(x) +2\sin'(x)(-\sin(x)) = 0 \end{eqnarray}$$ hence $$\sin^2(x) + \cos^2(x)$$ is constant, since its derivative is 0. Plugging in $$x=0$$ we see that it must equal 1.

Proof using Euler's formula: $$\sin^2(x) + \cos^2(x) = \left(\cos x + i \sin x\right)\left(\cos x - i \sin x\right) =\left(\cos x + i \sin x\right)\left(\cos (-x) + i \sin (-x)\right)= e^{i x} e^{-i x} = 1$$ Proof using Taylor series: Using the Taylor series:$$\begin{eqnarray} \sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\\ \cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \end{eqnarray}$$ we have $$\sin^2(x) = \left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\right)^2 = \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m} x^{2n + 2m + 2}}{(2n+1)!(2m+1)!} = \sum_{k=0}^\infty x^{2k+2}(-1)^k \sum_{j=0}^k \frac{1}{(2j+1)!(2(k-j)+1)!}= \sum_{k=0}^\infty \frac{x^{2k+2}(-1)^k}{(2k+2)!}\sum_{j=0}^k \binom{2k+2}{2j+1} = -\sum_{k=1}^\infty \frac{x^{2k}(-1)^k}{(2k)!}\sum_{j=0}^{k-1} \binom{2k}{2j+1}$$ and $$\cos^2(x) =\left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}\right)^2 = \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m} x^{2n + 2m}}{(2n)!(2m)!} = \sum_{k=0}^\infty x^{2k}(-1)^k \sum_{j=0}^k \frac{1}{(2j)!(2(k-j))!} = \sum_{k=0}^\infty \frac{x^{2k}(-1)^k}{(2k)!}\sum_{j=0}^k \binom{2k}{2j} = 1 + \sum_{k=1}^\infty \frac{x^{2k}(-1)^k}{(2k)!}\sum_{j=0}^k \binom{2k}{2j}$$ Adding these together:$$\sin^2(x) + \cos^2(x) = 1+\sum_{k=1}^\infty\frac{x^{2k}(-1)^k}{(2k)!}\left(\sum_{j=0}^k \binom{2k}{2j}-\sum_{j=0}^{k-1} \binom{2k}{2j+1}\right)=1+\sum_{k=1}^\infty\frac{x^{2k}(-1)^k}{(2k)!}\left(\sum_{j=0}^{2k} (-1)^j\binom{2k}{j}\right) = 1+\sum_{k=1}^\infty\frac{x^{2k}(-1)^l}{(2k)!} (1 + (-1))^{2k} = 1$$ We use the binomial theorem on the second last step.

Carlson's Theorem: Major overkill to use this one, but it still works. Notice $$\cos(x)$$, and $$\sin(x)$$ have exponential type 1. Thus $$f(x) = \cos^2(\frac{\pi}{4}x) + \sin^2(\frac{\pi}{4}x) - 1$$ has exponential type at most $$\frac{\pi}{2} < \pi$$. By inspection, you can see that $$f(x) = 0$$ for $$x\in \mathbb{N}$$. Hence Carlson's theorem implies $$f(x) = 0$$ identically.

• It's called the "cosine angle sum" formula, although I'd argue the way it's used is pretty angelic. :) – Lieutenant Zipp Apr 26 at 18:29

Calculus

Let $$f(x)=\sin^2x + \cos^2x$$. Then $$f'(x)=0$$ and so $$f$$ is constant, $$f(x)=f(0)=1$$.

This proof depends on $$\sin'=\cos$$, $$\cos'=-\sin$$.

• (+1) Nice approach :) – Learning Apr 26 at 15:14

A few proofs I came up with are:

Proof 1: This uses real analysis. Let $$f(x) = \sin^2x + \cos^2x \implies f'(x) = 2\sin x \cos x - 2\sin x \cos x = 0$$. Thus, $$f(x)$$ is a constant function. To find the value of $$f(x)$$, it is sufficient to find $$f(c)$$, where c is any convenient constant (say 0). Therefore, $$f(x) = f(0) = 1$$

Proof 2: This uses Euler's representation of complex numbers. We can represent a complex number in the form $$z = re^{ix}$$, where $$r$$ is the modulus and $$z$$ is the argument. If we take the number $$z = e^{ix} = \cos x + i\sin x$$ (De Moivre's theorem), then $$|z| = 1 = \sqrt{\cos^2x + \sin^2x} \implies \cos^2x + \sin^2x=1$$.

One way is to use the complex definitions of sine and cosine. $$\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i} \\\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$$ \begin{align*} \sin^{2}\theta + \cos^{2}\theta &= \Big( \frac{e^{i\theta}-e^{-i\theta}}{2i}\Big)^{2}+\Big(\frac{e^{i\theta}+e^{-i\theta}}{2} \Big)^{2} \\ &= -\frac{e^{2i\theta}-2e^{i\theta}e^{-i\theta}+e^{-2i\theta}}{4}+\frac{e^{2i\theta}+2e^{i\theta}e^{-i\theta}+e^{-2i\theta}}{4} \\ &=-\frac{e^{2i\theta}-2+e^{-2i\theta}}{4}+\frac{e^{2i\theta}+2+e^{-2i\theta}}{4} \\ &=\frac{4}{4} \\&=1\end{align*}

$$\left(\frac{e^{ix}+e^{-ix}}{2}\right)^2+\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2$$ $$=\frac{1}{4}\left[\left(e^{2ix}+2+e^{-2ix}\right)-\left(e^{2ix}+2+e^{-2ix}\right)\right]=\frac{4}{4}=1$$

• obtain expressions for $\sin(x)$ and $\cos(x)$ using a manipulation of Euler's formula – Henry Lee Apr 26 at 14:56

1) With identity $$a^2-b^2= (a-b)(a+b)$$,

$$\cos^2x+ \sin^2x = (\cos x-i\sin x)(\cos x+i\sin x)=e^{-i x}e^{i x}=1$$

2) With double angle identities,

$$\cos^2x+ \sin^2x =\frac12 (1+\cos 2 x)+\frac12(1-\cos 2x)=1$$

3) with half-angle subs $$t= \tan\frac x2$$,

$$\cos^2x+ \sin^2x =\left(\frac{1-t^2}{1+t^2}\right)^2 +\left(\frac{2t}{1+t^2}\right)^2 =\left(\frac{1+t^2}{1+t^2}\right)^2 =1$$

4) As an integral,

$$f(x)=\cos^2x+ \sin^2x =f(0) + \int_0^x f’(t)dt = 1+0=1$$

5) With $$x=it$$, apply $$\sin(it)= i\sinh t$$ and $$\cos(it)=\cosh t$$

$$\cos^2x+ \sin^2x = \cosh^2 t - \sinh^2 t = \left(\frac{e^t+e^{-t}}{2}\right)^2 -\left(\frac{e^t-e^{-t}}{2}\right)^2 =1$$

Special case of the addition theorems $$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$$ and $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$ by letting $$a=-b=x$$ in the second theorem. (The slickest prove of those theorems undoubtedly comes from Erhard Schmidt, see Proof of the angle sum identity for $\sin$.)

• It's not clear how you're using the sine formula to get $\cos^2(x) + \sin^2(x) = 1$. (Letting $a=-b=x$ yields the trivial $\sin(0) = \sin(x)\cos(x) - \sin(x)\cos(x)$) – Dark Malthorp Apr 28 at 18:35
• Only the theorem for $\cos$ is needed: $$1=\cos(0)=\cos(x)\cos(-x)-\sin(x)\sin(-x)=\cos^2(x)+\sin^2(x).$$ – Michael Hoppe Apr 29 at 7:50