How do you prove the following trigonometric identity: $$ \sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
[Math] Prove $\sin^2\theta + \cos^2\theta = 1$
trigonometry
Related Solutions
Here is an alternative using exponential forms:
$$ \begin{align*} 1+\tan^2 \theta &=1+\left( \frac{e^{i \theta}-e^{-i \theta}}{i\left( e^{i \theta}+e^{-i\theta} \right)} \right)^2 \\ &=1-\frac{\left( e^{i \theta}-e^{-i \theta} \right)^2}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\ &=\frac{\left( e^{i \theta}+e^{-i\theta} \right)^2-\left( e^{i \theta}-e^{-i \theta} \right)^2}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\ &=\frac{e^{2i\theta}+2+e^{-2i\theta}-e^{2i\theta}+2-e^{-2i\theta}}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\ &=\frac{4}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\ &=\left( \frac{2}{ e^{i \theta}+e^{-i\theta}} \right)^2 \\ &= \sec^2 \theta. \end{align*} $$
Here is an entirely different approach that focuses on the geometry of a right triangle.
Form a right triangle with angle $\theta$. Let $y$ be the side opposite $\theta$, $x$ be the side adjacent $\theta$, and label the hypotenuse $r$, where $r^2=x^2+y^2$ (by theorem of Pythagoras).
We can read trigonometric definitions right from the triangle as corresponding ratios of sides. Specifically, for angle $\theta$, $\tan \theta = \dfrac{y}{x}$, and $\sec \theta = \dfrac{r}{x}$. We can now write,
$$ \begin{align*} 1+\tan^2 \theta &= 1+\left( \frac{y}{x} \right)^2 \\ &=1+\frac{y^2}{x^2} \\ &=\frac{x^2+y^2}{x^2} \\ &=\frac{r^2}{x^2} \\ &= \left(\frac{r}{x}\right)^2 \\ &=\sec^2 \theta. \end{align*} $$
Here we repeat an idea used in the question Prove $\sin^2\theta+\cos^2\theta=1$ but it's slightly different since the functions $\cot$ and $\csc$ aren't defined on $\mathbb R$.
Let $$f(\theta)=\csc^2\theta-\cot^2\theta$$ then $f$ is defined on $\mathbb R\setminus\{k\pi,\; k\in\mathbb Z\}$ and we verify that $f'(\theta)=0$ so $f$ is constant in every interval $(k\pi,(k+1)\pi)$ and we conclude the result from the equality $$f\left(k\pi+\frac{\pi}{2}\right)=1$$
Best Answer
Let me contribute by this so let $$f(\theta)=\cos^2\theta+\sin^2\theta$$ then it's simple to see that $$f'(\theta)=0$$ then $$f(\theta)=f(0)=1$$