# $\sin^2(x)+\cos^2(x) = 1$ using power series

In an example I had to prove that $\sin^2(x)+\cos^2(x)=1$ which is fairly easy using the unit circle. My teacher then asked me to show the same thing using the following power series:$$\sin(x)=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}$$ and $$\cos(x)=\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k)!}$$ However, if I now take the squares of these values I get a really messy result that I can't simplify to 1.

Could anyone give me a hint on how to deal with this question?

• Have you tried just trying to find the power series manually (like find the first term, then the second and so on) and see if you can construct a power series)? Mar 26, 2014 at 21:59
• Good idea, I have only focussed on the power series given in the example. Will try to construct a joint power series tomorrow. Mar 26, 2014 at 22:02
• Hint: Differentiate the power series. We get that (surprise) the derivative of sine is cos, and the derivative of cos is $\dots$. Let $f(x)=\sin^2 x+\cos^2 x$. Show its derivative is $0$. Mar 26, 2014 at 22:05

Hint: Differentiate $\cos^2x+\sin^2x$, and notice that it is $2\cos x\cos'x+2\sin x\sin'x$. Evaluate the expressions of $\sin'x$ and $\cos'x$, and notice that they are the same as those of $\cos x$ and $-\sin x$ respectively. Then you get $\big(\cos^2x+\sin^2x\big)'=0\iff\cos^2x+\sin^2x=$ constant. How do we guess which constant exactly ? By computing $\sin^20+\cos^20$.

Hint: Differentiate the power series term by term. We get that (surprise!) the derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$. Let $f(x)=\sin^2 x+\cos^2 x$. Show its derivative is $0$. Thus $f(x)$ is constant.

It is not necessary to use derivatives. It is much simpler if we connect the series for $\sin x$ and $\cos x$ with the series of $\exp(x)$. From the given series it is easy to show that $\cos x + i\sin x = \exp(ix)$ and $\cos x - i\sin x = \exp(-ix)$ where $$\exp(z) = \sum_{n = 0}^{\infty}\frac{z^{n}}{n!}$$ and using binomial theorem it is easy to show that series of $\exp(z)$ satisfies $$\exp(z + z') = \exp(z)\cdot \exp(z')$$ and therefore \begin{aligned}\cos^{2}x + \sin^{2}x &= (\cos x + i\sin x)(\cos x - i\sin x)\\&= \exp(ix)\exp(-ix)\\&= \exp(0)\\&=1\end{aligned}

You can also prove this identity directly from the power series \begin{align} \cos x &= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n)!} x^{2n},\\ \sin x &= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)!} x^{2n + 1}. \end{align}

The following is modified from the discussion on Wikipedia's article on the Pythagorean Trigonometric Identity.

Squaring each of these series using the Cauchy Product $$\left(\sum_{i=0}^\infty a_i x^i\right) \cdot \left(\sum_{j=0}^\infty b_j x^j\right) = \sum_{k=0}^\infty \left(\sum_{l=0}^k a_l b_{k-l}\right) x^k\,,$$ and combining the factorials into a binomial coefficient we get \begin{align} \cos^2 x & = \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i)!} \frac{(-1)^j}{(2j)!} x^{(2i) + (2j)} \\ & = \sum_{n = 0}^\infty \left(\sum_{i = 0}^n \frac{(-1)^n}{(2i)!(2(n - i))!}\right) x^{2n} \\ & = 1 + \sum_{n = 1}^\infty \left( \sum_{i = 0}^n {2n \choose 2i} \right) \frac{(-1)^n}{(2n)!} x^{2n}\,,\\ \sin^2 x & = \sum_{i = 0}^\infty \sum_{j = 0}^\infty \frac{(-1)^i}{(2i + 1)!} \frac{(-1)^j}{(2j + 1)!} x^{(2i + 1) + (2j + 1)} \\ & = \sum_{n = 1}^\infty \left(\sum_{i = 0}^{n - 1} \frac{(-1)^{n - 1}}{(2i + 1)!(2(n - i - 1) + 1)!}\right) x^{2n} \\ & = \sum_{n = 1}^\infty \left( \sum_{i = 0}^{n - 1} {2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n}. \end{align}

Adding the squared series we can combine the odd and even terms then use the binomial theorem to simplify the internal sum to zero: \begin{align} \cos^2 x + \sin^2 x & = 1 + \sum_{n = 1}^\infty \left(\sum_{i = 0}^{n}{2n \choose 2i} - \sum_{i = 0}^{n - 1}{2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n} \\ & = 1 + \sum_{n = 1}^\infty \left(\sum_{j = 0}^{2n}(-1)^j{2n \choose j} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n} \\ & = 1 + \sum_{n = 1}^\infty \left(1-1\right)^{2n} \frac{(-1)^{n - 1}}{(2n)!} x^{2n} = 1\,. \end{align}

• I fixed a typo for you 😄 Apr 3, 2020 at 18:24

There is an easy solution using complex numbers (but not the complex exponential): \begin{align} \sin^2 x + \cos^2 x & = \left(\cos x + i \sin x\right)\left(\cos x - i \sin x\right) \\ &=\left(\sum_{j=0}^{\infty} (-1)^j \frac{x^{2j}}{(2j)!} + i \sum_{j=0}^{\infty} (-1)^j \frac{x^{2j+1}}{(2j+1)!}\right)\left(\sum_{j=0}^{\infty} (-1)^j \frac{x^{2j}}{(2j)!} - i \sum_{j=0}^{\infty} (-1)^j \frac{x^{2j+1}}{(2j+1)!}\right)\\ &=\left(\sum_{k=0}^{\infty} i^k \frac{x^{k}}{k!}\right) \cdot \left(\sum_{k=0}^{\infty} (-i)^k \frac{x^{k}}{k!}\right). \\ \end{align} (Of course these last two factors are equal to $$e^{ix}$$ and $$e^{-ix}$$, but we do not need this fact: we are just manipulating power series here.) We need to show that the above product is $$1$$, so that the coefficient of $$x^\ell$$ in the above product is $$1$$ if $$\ell=0$$ and $$0$$ otherwise.

Expanding the product, we get that the coefficient of $$x^\ell$$ equals $$\sum_{k_1+k_2=\ell} \frac{i^{k_1}}{k_1!} \cdot \frac{(-i)^{k_2}}{k_2!}=\sum_{k_1+k_2=\ell} (-1)^{k_2} \frac{i^{\ell}}{k_1!k_2!} = \sum_{k_1=0}^\ell (-1)^{\ell-k_1}\frac{i^{\ell}}{\ell!} \binom{\ell}{k_1}.$$ In view of the well-known binomial coefficient identity $$\sum_{k=0}^{\ell} (-1)^k \binom{\ell}{k}=0$$ for any integer $$\ell>0$$, which may be obtained by expanding the identity $$(1-1)^{\ell}=0$$, the above sum vanishes when $$\ell>0$$, and when $$\ell=0$$ it equals $$1$$, as can be computed directly. Which is precisely what we had to show.

Define $$f: \mathbb{R} \rightarrow \mathbb{R}$$ by $$f(x) = \text{sin}^2(x) + \text{cos}^2(x).$$ From the chain rule it follows that $$f'(x) = 2\cdot \text{sin}(x) \cdot \text{sin'}(x) + 2\cdot \text{cos}(x) \cdot \text{cos'}(x) = 2\cdot \text{sin}(x) \cdot \text{cos}(x) + 2\cdot \text{cos}(x) \cdot (-\text{sin}(x)) = 0,$$ as we know that $$\text{sin'}(x) = \text{cos}(x)$$ and $$\text{cos'}(x) = -\text{sin}(x)$$. We will now show that if the derivative equals zero, the function is constant, using the Mean-value theorem. Let $$a,b\in \mathbb{R}$$, with $$a. As $$f$$ is differentiable, and hence also continuous, on $$\mathbb{R}$$, we know there exists a $$c\in (a,b)$$ satisfying $$f'(c) = \frac{f(b)-f(a)}{b-a}.$$ As we know that $$f'(c) = 0$$ for all $$c \in \mathbb{R}$$, multiplying both sides of the equation by $$b-a$$ yields $$f(b)-f(a)=0$$. Equivalently, $$f(b) = f(a)$$ for all $$a,b\in \mathbb{R}$$, where $$a. Hence $$f$$ is constant on $$(a,b)$$. Note that $$f(0)= \text{sin}^2(0) + \text{cos}^2(0) = ( \sum_{k=0}^\infty (-1)^k \cdot \frac{1}{(2k+1)!} 0^{2k+1} )^2 + ( \sum_{k=0}^\infty (-1)^k \cdot \frac{1}{(2k)!} 0^{2k} )^2 = 0 + 1 = 1,$$ as only the first term of the cosine power series yields a nonzero value, namely 1 ($$(-1)^0 \cdot \frac{1}{(0)!}0^0=1$$, since $$(-1)^0=1$$, $$0!=1$$ and $$0^0=1$$).

In conclusion, for all $$x \in \mathbb{R}$$, $$f(x)=1$$, so that $$\sin ^{2}(x)+\cos ^{2}(x)=1.$$