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$ y = 4x^4 - 2x^2 + 5x + 7 $

$ -\frac{5}{3}x^{-\frac{2}{3}} + 2\sec^2 x + \frac{1}{2\sqrt{x}} $

$ \frac{2 - x}{(x + 1)(x^2 + 2)} = \frac{1}{x + 1} - \frac{x}{x^2 + 2} $

$ 0 \leq x \leq \pi \Rightarrow \sin x \geq 0 $

$ \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} $

$ \sin 20^{\circ} + \sin 30^{\circ} = 2\sin\left(\frac{20^{\circ} + 30^{\circ}}{2}\right)\cos\left(\frac{20^{\circ} - 30^{\circ}}{2}\right) $

$ \left[(6\cot x - 5) \cdot \frac{\sec^2 x}{\tan x}\right] \cdot \frac{\sin^3 x}{\cos(2x)} $

$ (1 + 0.1)^3 \approx 1 + 3 \times 0.1 + \frac{3(3 - 1)}{2!}(0.1)^2 + \frac{3(3 - 1)(3 - 2)}{3!}(0.1)^3 + \cdots $

$ f(1) = f(2) = 1,\ f(n + 2) = f(n) + f(n + 1);\ n = 1,2,3,\ldots $

The inverse sine function can be denoted as $\sin^{-1}x$ or $\arcsin x$, and the domain of $x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

The derivative of $\tan x$ is $\sec^2 x$ as $ f’(x) = \frac{d}{dx}(\tan x) = \sec^2 x $

$ \displaystyle\int_{1}^{2} \frac{1}{2\sqrt{x}} dx = \sqrt{2} - 1 $

$ \lim\limits_{h \to 0} \frac{\sin h}{h} = 1 $

$ \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} $

$ z_1 \times z_2 = 4 + 5i $

$ (1 + 2i) \cdot (2 - 3i) \cdot (5 + 8i) $

$ \frac{z_1}{z_2} = \frac{r_1}{r_2} \left( \cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2) \right) $

$ \left| \frac{\overline{z_1 + z_2}}{z_3 - z_4} \right| $

$ \cosec x $ (Hint: I use \textbackslash newcommand to create user-defined command)

$ \left( \frac{a}{b} \right)^n = \frac{a^n}{b^n} $

$ a^{m/n} = a^{\frac{m}{n}} = \sqrt[n]{a^m} = \left( \sqrt[n]{a} \right)^m $

$ \sin^2 {x}+ \cos^2 {x}+ \tan^2 {x} = 0 $

$ (f \circ g)(x) $ is a composite function with respect to $x$

$ x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-b \pm \sqrt{b^2 \pm 4ac}}{2a} \quad (a \neq 0) $

$ a^x = y \ \Leftrightarrow \ x = \log_{a} y $

$ \log_{a} xy - \log_{a}\left( \frac{x}{y} \right) = \log_{a} y^2 $

$ \log_{y} x = \frac{\log_{a} x}{\log_{a} y} $

$ \frac{1}{3}\ln\left( \frac{x^2}{x - 1} \right) + 2\log_{10} x - \log_{x}(x + 1) $, where $x > 1$

$ \sum n^3 = \frac{n^2(n + 1)^2}{4} $

The harmonic series is shown as $ \sum\limits_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots $