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.
$\endgroup$ 28 Answers
$\begingroup$Here's an alternative hint with a more geometrical flavour: what are the angles in an isosceles right-angled triangle?
$\endgroup$ $\begingroup$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 $$ \sin(x) - \cos(x) = \sqrt{2}\sin(x - 45^\circ) $$
$\endgroup$ 7 $\begingroup$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).
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ 0 $\begingroup$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 $\tan x=1$.
$\endgroup$ $\begingroup$For posterity, here's a unit circle interpretation of the equation $\cos\theta = \sin\theta$, viewing the circle as parametrized by $$ x = \cos\theta,\qquad y = \sin\theta. $$
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 $\cos x$): $$\cos t=\frac{e^{it}+e^{-it}}2=\frac{e^{it}-e^{-it}}{2i}=\sin t$$ $$(1+i)e^{-it}=(1-i)e^{it}$$ $$e^{2it}=\frac{1+i}{1-i}=i$$ $$\cos2t+i\sin2t=i$$ $$\implies\cos 2t=0 \text{ }\cap \sin2t=1$$ $$x=\frac{\pi}4+n\pi,n\in\Bbb{N}$$
$\endgroup$ $\begingroup$$cos(x)=sin(x) \implies x = \pi n-\frac{3\pi}{4}$ where n is an integer.
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