% more Blackboard letters
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\def\bH{\mbb{H}}

% brackets & parens stuff
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\newcommand{\sbr}[1]{\left[#1\right]}
\newcommand{\brc}[1]{\left\{#1\right\}}
\newcommand{\lbr}[1]{\left\langle#1\right\rangle}

% thanks Tuong for the \fxn cmd
\newcommand{\fxn}[5]{#1:\begin{array}{rcl}#2&\longrightarrow & #3\\[-0.5mm]#4&\longmapsto &#5\end{array}}
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\renewcommand{\L}{\mathcal{L}}
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% more function names
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\DeclareMathOperator{\arctanh}{arctanh}
\DeclareMathOperator{\arccsch}{arccsch}
\DeclareMathOperator{\arcsech}{arcsech}
\DeclareMathOperator{\arccoth}{arccoth}
\DeclareMathOperator{\cis}{cis}
\DeclareMathOperator{\sinc}{sinc}
\DeclareMathOperator{\sgn}{sgn}
\DeclareMathOperator{\rect}{rect}
\DeclareMathOperator{\tri}{tri}

% linalg stuff
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\renewcommand{\j}{\hat{\jmath}}
\renewcommand{\k}{\hat{k}}
\renewcommand{\l}{\hat{\ell}}
\DeclareMathOperator{\cof}{cof}
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% for fun
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\newcommand{\meow}{\approx\mkern-5mu\mathcal{O}\omega\mathcal{O}\mkern-5mu\approx}
\newcommand{\limdefn}{f'(x)=\lim_{h\to0}\tfrac{f(x+h)-f(x)}h}
\newcommand{\trig}{\text{A relatively small list of trigonometric identities:}\\\\\text{defining relations:}\\\csc(x)=\tfrac1{\sin(x)}\quad\tan(x)=\tfrac{\sin(x)}{\cos(x)}\\\sec(x)=\tfrac1{\cos(x)}\quad\cot(x)=\tfrac1{\tan(x)}=\tfrac{\cos(x)}{\sin(x)}\\\\\text{pythagoras:}\\\sin^2(x)+\cos^2(x)=1\\1+\cot^2(x)=\csc^2(x)\\\tan^2(x)+1=\sec^2(x)\\\\\text{cofunction, negative angle, and supplement:}\\\sin(x)=\cos(\tfrac\pi2-x)\quad\sin(-x)=-\sin(x)\\\cos(x)=\sin(\tfrac\pi2-x)\quad\cos(-x)=\cos(x)\\\tan(x)=\cot(\tfrac\pi2-x)\quad\tan(-x)=-\tan(x)\\\sin(\pi-x)=\sin(x)\\\cos(\pi-x)=-\cos(x)\\\tan(\pi-x)=-\tan(x)\\\\\text{periodicity - for all }n\in\bZ\text{, we have:}\\\sin(x+2\pi n)=\sin(x)\\\cos(x+2\pi n)=\cos(x)\\\tan(x+\pi n)=\tan(x)\\\\\text{addition:}\\\sin(x\pm y)=\sin(x)\cos(y)\pm\cos(x)\sin(y)\\\cos(x\pm y)=\cos(x)\cos(y)\mp\sin(x)\sin(y)\\\tan(x\pm y)=\tfrac{\tan(x)\pm\tan(y)}{1\mp\tan(x)\tan(y)}\\\\\text{double angle:}\\\sin(2x)=2\sin(x)\cos(x)\\\cos(2x)=\cos^2(x)-\sin^2(x)\\\cos(2x)=2\cos^2(x)-1=1-2\sin^2(x)\\\tan(2x)=\tfrac{2\tan(x)}{1-\tan^2(x)}\\\\\text{sum-to-product and product-to-sum:}\\\cos(x)+\cos(y)=2\cos(\tfrac{x+y}2)\cos(\tfrac{x-y}2)\\\cos(x)-\cos(y)=-2\sin(\tfrac{x+y}2)\sin(\tfrac{x-y}2)\\\sin(x)\pm\sin(y)=2\sin(\tfrac{x\pm y}2)\cos(\tfrac{x\mp y}2)\\2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y)\\2\sin(x)\cos(y)=\sin(x+y)+\sin(x-y)\\2\cos(x)\cos(y)=\cos(x+y)+\cos(x-y)\\\\\text{Hope you find whatever identities you need!}}