How to evaluate $\arctan(1)$ I thought 
$$\arctan(1) = \dfrac{\arcsin(1)}{\arccos(1)}$$
Sin hits 1 at $\pi/2$, and Cos hits 1 at $0$ and $2\pi$
So $\dfrac{\arcsin(1)}{\arccos(1)} = \dfrac{1}{4}$
But the solution says it is $\pi/4$...
 A: $\arctan(x)\neq \frac{\arcsin(x)}{\arccos(x)}$
$\arctan(1)=\frac{\pi}{4}$ because $\tan(\frac{\pi}{4})=1$. The latter you can see in a so called "45-45-90" triangle.
A: By (one) definition, the arctangent of $1$ is the unique number in the interval $(-\pi/2,\pi/2)$ whose tangent is $1.$ I suspect that you already know that $\tan x=\frac{\sin x}{\cos x}$ for all real numbers $x.$ Consequently, $\arctan 1$ is the unique number in the interval $(-\pi/2,\pi/2)$ whose sine and cosine are equal. Depending on your definition of sine and cosine, it may be more or less obvious that $$\arctan(1)=\frac\pi4.$$
A: You have
$$
\tan \left( \frac\pi4\right)=\frac{\sin\left( \frac\pi4\right)}{\cos\left( \frac\pi4\right)} = \frac{\frac{\sqrt{2}}2}{\frac{\sqrt{2}}2}=1
$$ giving
$$
\frac\pi4=\arctan (1).
$$
A: In order for the inverse of cosine to be defined, it must be restricted to a domain for which it is one-to-one. Therefore, we restrict the domain of $\cos$ to be from $-\pi$ to $\pi$. So the range of $\cos^{-1} $ is as follows.
$$-\pi \leq \cos^{-1}(y) \leq \pi$$.
So $2\pi \neq cos^{-1}(1)$. Indeed, $\cos^{-1}(1) = 0$, making the equation
$$\frac{\sin^{-1}(1)}{\cos^{-1}(1)} = \frac{\pi/2}{0}$$
undefined. 
To find $\tan^{-1}(1)$, we need to find $x$ so that $\tan(x) =1$. Thus,
$$\tan(x) = \frac{\sin(x)}{\cos(x)} = 1$$  
implies that $\sin(x) = \cos(x)$. So $x$ must be $k \pi /4$ where $k$ is any integer. But as we are trying to solve for $x$ so that $\tan^{-1}(1) = x$, $x$ has to be in the range of $-\pi$ to $\pi$, i.e., $x = \pi/4$.
A: You are not stupid. You made an adventurous generalisation hoping that a new functional relation F should still respect the earlier ones.
$$\frac{  f(x)}{g(x) }=\frac{ F( f(x))} {F(g(x) )} $$ 
Unfortunately not true, in general it cannot be brought on to a new stage still respecting earlier functional relationships.If F is multiplication by a constant, then of course it is OK.
