show $\sin(x)$ and $\tan(x)$ are increasing Show that the functions sin and tan are each increasing on $(-π/2, π/2)$. Hence define the functions $\sin^{-1}$ and $\tan^{-1}$ (on $(-1,1)$ and $\Bbb R$ respectively), prove them differentiable, and compute their derivatives. 
 A: Hint: $ \sin x - \sin y = 2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2}) > 0 $ when $x-y>0$.
A: David H's hint is a good one.
If $x,y \in (-\pi/2,\pi/2)$ then $|x-y|<\pi$ therefore $\displaystyle \frac{|x-y|}{2}<\pi/2$. 
So, if we suppose $x>y$ then $\displaystyle 0<\frac{x-y}{2}<\frac{\pi}{2}$
Also, $|x+y|<\pi$ and if we suppose $x>y$ then $\displaystyle -\frac{\pi}{2}<\frac{x-y}{2}<\frac{\pi}{2} $.
But $\sin(x)$ is positive for $x \in (0,\pi/2)$ and $\cos(x)$ is positive for $x \in (-\pi/2,\pi/2)$ therefore by the trigonometric identity $\displaystyle \sin(x)-\sin(y)=2 \cos(\frac{x+y}{2})\sin(\frac{x-y}{2})$ and using the fact that $\displaystyle \cos(\frac{x+y}{2})>0$ and  $\displaystyle \sin(\frac{x-y}{2})>0$ we see that $\sin(x)-\sin(y)>0$.
To prove that $\sin(x)$ is differentiable, we must show that the limit
$$ \lim_{\Delta{x} \to 0} \frac{\sin(x+\Delta{x})-\sin(x)}{\Delta{x}}$$
exists for every $x \in \mathbb{R}$.
You're going to need to know that 
$$ \lim_{x \to 0} \frac{\sin(x)}{x}=1$$
and also
$$ \lim_{x \to 0} \frac{1-\cos(x)}{x}=0$$
Now just expand $\sin(x+\Delta{x})=\sin(x)\cos(\Delta{x})+\cos(x)\sin(\Delta{x})$:
$$ \lim_{\Delta{x} \to 0} \frac{\sin(x)\cos(\Delta{x})+\cos(x)\sin(\Delta{x})-\sin(x)}{\Delta{x}}$$
$$= \lim_{\Delta{x} \to 0} \frac{\sin(x)(\cos(\Delta{x})-1)+\cos(x)\sin(\Delta{x})}{\Delta{x}}$$
$$=\lim_{\Delta{x} \to 0}\sin(x)\cdot \frac{\cos(\Delta{x})-1}{\Delta{x}}+\lim_{\Delta{x} \to 0}\cos(x) \cdot \frac{\sin(\Delta{x})}{\Delta{x}}$$
$$= 0 + \cos(x).1 = \cos(x)$$
since $\cos(x)$ is well-defined for all $x \in \mathbb{R}$ we'll see that $\sin(x)$ is differentiable over the real line and its derivative is equal to $\cos(x)$.
You can do the same with $\cos(x)$ and use the identity $$\cos(x+\Delta{x})=\cos(x)\cos(\Delta{x})-\sin(x)\sin(\Delta{x})$$ to expand $\cos(x+\Delta{x})$ in the limit to get $(\cos(x))'=-\sin(x)$.I leave it to you as an exercise. The proof uses the same techniques I used for $(\sin(x))'=\cos(x)$
A: Look, the algebra and the calculus is all true and wonderful. But why don't we look at the very basis here? What is the definition of a sine? The definition (not a theorem!) is that it is the vertical coordinate of a point P on the unit circle, as a function of the angle, determined by OP with the pos. x-axis. This is exactly how Hipparchus, arguably the founding father of trigonometry introduced the ratios which we now call sine and cosine. There was a need for a relation between polar coordinates and rectangular coordinates to study things like planetary movements, including moon and sun. Now follow P from the angles listed in the question and what happens with the y-coordinate? It increases. The calculus confirms this very basic property of a y-value of a point traveling from Quadrant 4 all the way up to Quadrant 1. Sometimes history approaches complicated problems in a very basic manner. It doesn't take away the fact that the calculus is a neat exercise, of course.
A: Since this is a calculus question and the derivatives of the inverses of these functions will be evaluated, I assume that we can note that
$$
\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)=\cos(x)\tag{1}
$$
and
$$
\frac{\mathrm{d}}{\mathrm{d}x}\tan(x)=\sec^2(x)\tag{2}
$$
and that both $\cos(x)$ and $\sec^2(x)$ are positive on $\left(-\frac\pi2,\frac\pi2\right)$. This means that $\sin(x)$ and $\tan(x)$ are increasing on $\left(-\frac\pi2,\frac\pi2\right)$ by the Mean Value Theorem.
Note that by the Chain Rule,
$$
1=\frac{\mathrm{d}}{\mathrm{d}x}x=\frac{\mathrm{d}}{\mathrm{d}x}f\left(f^{-1}(x)\right)=f'\left(f^{-1}(x)\right)\frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)\tag{3}
$$
therefore,
$$
\frac{\mathrm{d}}{\mathrm{d}x}f^{-1}(x)=\frac1{f'\left(f^{-1}(x)\right)}\tag{4}
$$
Now use $(1)$ and $(2)$ with $(4)$ to get the derivatives of $\sin^{-1}(x)$ and $\tan^{-1}(x)$.
