8

We know: $$\sin{x} = x - \frac{x^3}{3!} + \frac{x^3}{5!}-\dotsb$$ and so on. Also, $$\cos{x} = 1 - \frac{x^2}{2!} +\frac{x^4}{4!}-\dotsb$$ and so on.

With the help of these expansions we need to prove that $\sin^2 x+\cos^2 x=1$ .

I tried generalizing $\sin{x}$ as $$\sum (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ and $\cos{x}$ as $$\sum (-1)^n\frac{x^{2n}}{(2n)!},$$ then squaring and adding. But it didn't get through. Please help!

Chappers
  • 67,606
User9523
  • 2,094

1 Answers1

4

I think the easiest way to see it is to notice that $e^{ix}=\cos(x)+i\sin(x)$ which is easily verified by writing down the Taylor expansion of both sides. Then

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

If you want to verify $f(y)=e^{y}e^{-y}=1$, here's how: first show $\frac{d}{dx}e^y=e^y$ from the Taylor series. Then show $f'(y)=f(y)-f(y)=0$ by the chain rule. Conclude $f(y)$ is a constant, and plug in $y=0$ to get $f(y)=1$.

Alex R.
  • 32,771
  • Since we are using the series based approach it is much easier to show that the series $g(z) = 1 + z + \dfrac{z^{2}}{2!} + \dfrac{z^{3}}{3!} + \cdots$ satisfies the property $g(z_{1} + z_{2}) = g(z_{1})g(z_{2})$ by directly multiplying the series and using binomial theorem. This easily gives $g(z)g(-z) = g(0) = 1$. – Paramanand Singh Apr 30 '15 at 11:52