When is $\sin(x) = \cos(x)$? How do I solve the following equation?
$$\cos(x) - \sin(x) = 0$$ 
I need to find the minimum and maximum of this function: 
$$f(x) = \frac{\tan(x)}{(1+\tan(x)^2}$$
I differentiated it, and in order to find the stationary points I need to put the numerator equal to zero. But I can't find a way to solve this trigonometric equation.
 A: 1)  $\cos^2 x + \sin^2 x = 1$
So $2 \cos^2 x = 1$
So $\cos x = \sin x = \pm \sqrt{\frac 12}$
2) $\sin x$ is the adjacent side of a right triangle.  $\cos x $ is the opposite side.  $\sin x = \cos x$ means the triangle is isoceles.  So the base angles are congruent.  So $x + x + 90 = 180$.
3) $\sin x = \cos (\frac {\pi}2 -x)$
so $\cos x = \sin x = \cos  (\frac {\pi}2-x)$
Can you figure it out now?
There are some quadrant issues to figure out but they aren't hard.
A: If you have an equation of the form
$$
\cos f(x)=\sin g(x)
$$
you can rewrite it as
$$
\cos f(x)=\cos\left(\frac{\pi}{2}-g(x)\right)
$$
and so
$$
f(x)=\frac{\pi}{2}-g(x)+2k\pi
\qquad\text{or}\qquad
f(x)=-\frac{\pi}{2}+g(x)+2k\pi
$$
In your particular case, $f(x)=g(x)=x$, so you have
$$
x=\frac{\pi}{2}-x+2k\pi
\qquad\text{or}\qquad
x=-\frac{\pi}{2}+x+2k\pi
$$
Of course, the second possibility gives no solution; the first case gives
$$
x=\frac{\pi}{4}+k\pi
$$

If your function is
$$
f(x)=\frac{\tan x}{(1+\tan x)^2}
$$
the derivative is
\begin{align}
f'(x)
&=\frac{(1+\tan^2x)(1+\tan x)^2-2(1+\tan x)(1+\tan^2x)\tan x}
       {(1+\tan x)^4}
\\[6px]
&=\frac{(1+\tan^2x)(1-\tan x)}{(1+\tan x)^3}
\end{align}
so it vanishes for $\tan x=1$.
A: Here's an alternative hint with a more geometrical flavour: what are the angles in an isosceles right-angled triangle?
A: Hint: Take the equation 
$$
\sin(x) = \cos(x)
$$
and divide both sides by $\cos(x)$ to get
$$
\tan(x) = 1
$$

Alternatively, using a sum-to-product formula, we can observe that
$$
\sin(x) - \cos(x) = \sqrt{2}\sin(x - 45^\circ)
$$
A: For posterity, here's a unit circle interpretation of the equation $\cos\theta = \sin\theta$, viewing the circle as parametrized by
$$
x = \cos\theta,\qquad
y = \sin\theta.
$$

A: Approach $1$ (Squaring):
$$(\sin x-\cos x)^2=0$$
$$(\sin^2x+\cos^2x)-2\sin x\cos x=0$$
$$1-\sin2x=0$$
$$\sin2x=1$$
$$2x=\frac{\pi}2+2n\pi,n\in\Bbb{N}$$
$$x=\frac{\pi}4+n\pi,n\in\Bbb{N}$$

Approach $2$ (By definition of $\sin x$ and $\cos x$):
$$\cos t=\frac{e^{it}+e^{-it}}2=\frac{e^{it}-e^{-it}}{2i}=\sin t$$
$$(1+i)e^{-it}=(1-i)e^{it}$$
$$e^{2it}=\frac{1+i}{1-i}=i$$
$$\cos2t+i\sin2t=i$$
$$\implies\cos 2t=0 \text{ }\cap \sin2t=1$$
$$x=\frac{\pi}4+n\pi,n\in\Bbb{N}$$
A: A direct approach: use the unit-circle definition of sine and cosine.
Consider a unit circle around the origin of a Cartesian plane.
Since in this problem $x$ is already in use as an angle,
we cannot label the two axes $x$ and $y$ as usual, so let's label them
$u$ (on the horizontal axis) and $v$ (on the vertical axis) instead.
Then the unit-circle definition says that if we take a point at an angle
$x$ radians counterclockwise around the unit circle, the coordinates
at that point will be
\begin{align}
u &= \cos x, \\
v &= \sin x.
\end{align}
The equation $\cos x = \sin x$ then tells us that $u = v$,
which is the equation of a line at $\frac\pi4$ radians ($45$ degrees)
through the origin.
But we got these coordinates in the first place as coordinates of a point
on the unit circle, so the solution must be at a point where the
line $u = v$ intersects the unit circle.
Draw a graph, as in the figure below: there are two such points.

The coordinates of these points happen to be
$\left(\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right)$
and $\left(-\frac{\sqrt2}{2},-\frac{\sqrt2}{2}\right)$,
but even without figuring that out, since you know the angle of the line
$u=v$ you can easily see the two possible values of
$x$ within the range $0$ to $2\pi$ radians ($0$ to $360$ degrees).
A: $cos(x)=sin(x) \implies x = \pi n-\frac{3\pi}{4}$ where n is an integer.
