Let $s(x)= x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \ $ and $ \ c(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots \ $ for real $x$.
How can we prove that $(s^2+c^2)'=0$ and $s^2+c^2=1$?
The fact that $s(x)=\sin x$ and $c(x)=\cos x$ can't be used.
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1 Answer
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1 Use the sum rule and chain rule to obtain $$\frac{d}{dx}(s(x))^2 + (c(x))^2) = 2s(x)s'(x) + 2c(x)c'(x).$$ Differentiate term by term to find that $s'(x) = c(x)$ and $c'(x) = -s(x)$. Hence $$\frac{d}{dx}(s(x))^2 + (c(x))^2) = 0.$$ Integrating both sides implies $$(s(x))^2 + (c(x))^2 = C,$$ where $C$ is some constant. Substitute $x = 0$ to conclude that $C = 1$.
- $\begingroup$ Got it! Thanks a lot! $\endgroup$J.doe– J.doe2016-04-15 01:58:33 +00:00Commented Apr 15, 2016 at 1:58