5.6 Identities

5.6.0.1 trig and Hyper trig identities

$$\begin{array}{rl}\cos \left(i\theta \right)& =\cosh \left(\theta \right)\\ \sin \left(i\theta \right)& =i\sinh \left(\theta \right)\end{array}$$

$$\begin{array}{rl}{\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)& =1\\ {\tan }^2\left(\theta \right)& =\frac{1}{{\cos }^2\left(\theta \right)}-1\\ & ={\sec }^2\left(\theta \right)-1\\ \frac{{\cos }^2\left(\theta \right)}{{\sin }^2\left(\theta \right)}+1& =\frac{1}{{\sin }^2\left(\theta \right)}\\ \frac{1}{{\tan }^2\left(\theta \right)}& =\frac{1}{{\sin }^2\left(\theta \right)}-1\\ {\cot }^2\left(\theta \right)& ={\csc }^2\left(\theta \right)-1\\ {\cosh }^2\left(\theta \right)-{\sinh }^2\left(\theta \right)& =1\end{array}$$

$$\begin{array}{rl}\sin \left(2\theta \right)& =2\sin \left(\theta \right)\cos \left(\theta \right)\\ \cos \left(2\theta \right)& ={\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right)\\ & =2{\cos }^2\left(\theta \right)-1\\ & =1-2{\sin }^2\left(\theta \right)\\ \tan \left(2\theta \right)& =\frac{2\tan \left(\theta \right)}{1-{\tan }^2\left(\theta \right)}\\ \sinh \left(2\theta \right)& =2\sinh \left(\theta \right)\cosh \left(\theta \right)\\ \cosh \left(2\theta \right)& =2{\cosh }^2\left(\theta \right)-1\\ \tanh \left(2\theta \right)& =\frac{2\tanh \left(\theta \right)}{1+{\tanh }^2\left(\theta \right)}\end{array}$$

$$\begin{array}{rl}\sin \left(\theta \right)& =\cos (\frac{\pi }{2}-\theta )\\ \cos \left(\theta \right)& =\sin (\frac{\pi }{2}-\theta )\end{array}$$

$$\begin{array}{rl}\sin (A+B)& =\sin A\cos B+\cos A\sin B\\ \sin (A-B)& =\sin A\cos B-\cos A\sin B\\ \cos (A+B)& =\cos A\cos B-\sin A\sin B\\ \cos (A-B)& =\cos A\cos B+\sin A\sin B\\ \tan (A+B)& =\frac{\tan A+\tan B}{1-\tan A\tan B}\\ \tan (A-B)& =\frac{\tan A+\tan B}{1+\tan A\tan B}\end{array}$$

$$\begin{array}{rl}{\sin }^2\left(\theta \right)& =\frac{1}{2}(1-\cos \left(2\theta \right))\\ {\cos }^2\left(\theta \right)& =\frac{1}{2}(1+\cos \left(2\theta \right))\\ {\tan }^2\left(\theta \right)& =\frac{1-\cos \left(2\theta \right)}{1+\cos \left(2\theta \right)}\end{array}$$

$$\begin{array}{rl}\sin A+\sin B& =2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\\ \sin A-\sin B& =2\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)\\ \cos A+\cos B& =2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\\ \cos A-\cos B& =-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\end{array}$$

$$\begin{array}{rl}\sin A\sin B& =\frac{1}{2}(\cos (A-B)-\cos (A+B))\\ \cos A\cos B& =\frac{1}{2}(\cos (A-B)+\cos (A+B))\\ \sin A\cos B& =\frac{1}{2}(\sin (A+B)+\sin (A-B))\\ \cos A\sin B& =\frac{1}{2}(\sin (A+B)-\sin (A-B))\end{array}$$

$$\begin{array}{rl}a\cos \left(\omega t\right)+b\sin \left(\omega t\right)& =A\sin (\omega t+\varphi )\\ & =A\cos (\omega t-\varphi )\\ A& =\sqrt{a^2+b^2}\\ \varphi & =\arctan \left(\frac{B}{A}\right)\\ \cos x+\sin x& =\sqrt{2}\sin (x+\frac{\pi }{4})\\ \cos x+\sin x& =\sqrt{2}\cos (x-\frac{\pi }{4})\end{array}$$

Laws of sines ($a,b,c$) are lengths of triangle sides and $A,B,C$ are facing angles.$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$ laws of cosine$$a^2=b^2+c^2-2bc\cos A$$

5.6.0.2 GAMMA function

$$\begin{array}{rl}\mathrm{\Gamma }(n)& =(n-1)!\\ \mathrm{\Gamma }(n+1)& =n(n-1)!\\ & =n\mathrm{\Gamma }(n)\end{array}$$

5.6.0.3 Sterling

For $n\gg 1$$$\mathrm{\Gamma }(n+1)=n!=\sqrt{2\pi }{n}^{n+\frac{1}{2}}{e}^{-n}$$