How to prove that $\frac{d}{dx} \left(\sin^2(x)+\cos^2(x)\right)=0$ I have seen in a mathematics.stackexchange.com thread that to prove that $\sin^2x+\cos^2x=1$ one have to prove that the derivative of $\left(\sin^2(x)+\cos^2(x)\right)$ is $0$.
But how can we prove it?
 A: $$
\begin{align}
\frac{d}{dx} (\sin x)^2 & = 2\sin x\cdot\frac{d}{dx} \sin x = 2\sin x\cos x \\[15pt]
\frac{d}{dx} (\cos x)^2 & = 2\cos x\cdot\frac{d}{dx}\cos x = 2\cos x\cdot(-\sin x)
\end{align}
$$
Now add those together.
A: Hint: $$\dfrac{\mathrm d}{\mathrm dx}\Big[\cos^2x+\sin^2x\Big]=\dfrac{\mathrm d}{\mathrm dx}\Big[\cos x\cos x\Big]+\dfrac{\mathrm d}{\mathrm dx}\Big[\sin x\sin x\Big].$$
Now use the product rule knowing that: $$\dfrac{\mathrm d}{\mathrm dx}\sin x=\cos x\quad\color{grey}{\text{and}}\quad\dfrac{\mathrm d}{\mathrm dx}\cos x=-\sin x.$$
A: First we can see easily that 
$$\frac{d}{dx} (\cos^2 x+\sin^2 x)=\frac{d}{dx}1$$
and since 
$$\frac{d}{dx} =0$$
Then we can plug it back into the original function to get
$$\frac{d}{dx} 1=0=\frac{d}{dx} (\cos^2 x+\sin^2 x)$$
But if you want to do this from scratch, we can also say that by the linearity of the derivative:
$$\frac{d}{dx} (\cos^2 x+\sin^2 x)=\frac{d}{dx} (\cos^2 x)+\frac{d}{dx} (\sin^2 x)$$ 
And now by the chain rule:
$$\frac{d}{dx} (\cos^2 x)+\frac{d}{dx} (\sin^2 x)=(2(\cos x)(-\sin x))+(2(\sin x)(\cos x))$$
This was the big step, now it's simple algebra:
$$(2(\cos x)(-\sin x))+(2(\sin x)(\cos x))=-2(\cos x)(\sin x)+2(\sin x)( \cos x)=0$$
and that is it! I hope this helps.
A: I am assuming only the trigonometric definitions of $\sin$ and $\cos$ functions.
 $\sin^2x+\cos^2x=1$ can be proved$^1$ by showing that $\dfrac{d}{dx}(\sin^2x+\cos^2x)=0$. This identity is equivalent to pythagorean theorem. Pythagorean theorem in turn can be proved with the use of calculus, e.g. see here($\angle EDC$ tends to $\angle BCA$ as $\Delta x$ tends to $0$). 

$^1$ By applying chain rule we can show that $\dfrac{d}{dx}f(x)=0$ where $f(x)=(\sin^2x+\cos^2x)$. We know if the derivative of a function is $0$ then the function is a constant.   
So $\dfrac{d}{dx}(\sin^2x+\cos^2x)=0 \implies (\sin^2x+\cos^2x)= constant\ \  \forall\ x$.
At $x=0$, $=(\sin^20+\cos^20)=(0+1)=1$.   
Now  we have computed the value of $(\sin^2x+\cos^2x)$ for a certain value of $x$(namely $x=0$), this value will remain same for every other value of $x$, since $(\sin^2x+\cos^2x)$ is a constant function.
  So $(\sin^2x+\cos^2x)=1\ \forall\ \ x$.
