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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?
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.
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: $$ \begin{align} \cos^2 x + \sin^2 x & = 1 + \sum_{n = 1}^\infty \left(\sum_{i = 0}^{n}{2n \choose 2i} - \sum_{i = 0}^{n - 1}{2n \choose 2i + 1} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n} \\ & = 1 + \sum_{n = 1}^\infty \left(\sum_{j = 0}^{2n}(-1)^j{2n \choose j} \right) \frac{(-1)^{n - 1}}{(2n)!} x^{2n} \\ & = 1 + \sum_{n = 1}^\infty \left(1-1\right)^{2n} \frac{(-1)^{n - 1}}{(2n)!} x^{2n} = 1\,. \end{align} $$
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 $$\begin{aligned}\cos^{2}x + \sin^{2}x &= (\cos x + i\sin x)(\cos x - i\sin x)\\&= \exp(ix)\exp(-ix)\\&= \exp(0)\\&=1\end{aligned}$$
