Trigonometric identities


\[\cases{{}\\ Cos\ -φ=Cos\ φ=Cos\left({π±φ}\right)\\ {}\\ Sin\ -φ=-Sin\ φ=Sin\left({π+φ}\right)\\ {}\\ \qquad \qquad \qquad Sin\ \mathbf{φ}=Sin\left({π-\mathbf{φ}}\right)\\ {}\\ Cos\ φ=Sin\left({φ+\frac π2}\right)\ ,\ Cos\frac π4=Sin\frac{3π}4=Sin\left({π-\mathbf{\frac π4}}\right)=Sin\mathbf{\frac π4}\\ {}\\ Sin\ Θ=Cos\left({Θ-\frac π2}\right)\ ,\ Sin\frac π4=Cos\left({-\frac π4}\right)=Cos\frac π4\\ {}} \]

\[\displaystyle{Sin\ x=\sum_{i=1}^n\left({-1}\right)^{n-1}\frac{x^{2n-1}}{\left({2n-1}\right)!}}\]

\[\displaystyle{Cos\ x=1-\sum_{i=1}^n\left({-1}\right)^{n-1}\frac{x^{2n}}{\left({2n}\right)!}}\]


 \[Sin\left({α±ß}\right)=Sin\ α·Cos\ ß±Cos\ α·Sin\ ß\]

 \[Cos\left({α±ß}\right)=Cos\ α·Cos\ ß∓Sin\ α·Sin\ ß\]

 \[Tan\left({α±ß}\right)=\frac{Tan\ α±Tan\ ß}{1∓Tan\ α·Tan\ ß}\]

 \[Cot\left({α±ß}\right)=\frac{Cot\ \mathbf{ß}·Cot\ \mathbf{α}∓1}{Cot\ \mathbf{ß}±Cot\ \mathbf{α}}\]


\[Sin\ α±Sin\ ß=2·Sin\frac{α±ß}2·Cos\frac{α∓ß}2\]

\[Cos\ α±Cos\ ß=2·{}^{Cos}_{Sin}\frac{α+ß}2·{}^{Cos}_{Sin}\frac{α-ß}2\]

\[Tan\ α±Tan\ ß=\frac{Sin\left({α±ß}\right)}{Cos\ α·Cos\ ß}\]

 \[Cot\ α±Cot\ ß=\frac{Sin\left({\mathbf{ß}±\mathbf{α}}\right)}{Sin\ \mathbf{ß}·Sin\ \mathbf{α}}\]


\[Sin^2\ α+Cos^2\ α=1=\\ =\left({Cos\ α-i·Sin\ α}\right)\left({Cos\ α+i·Sin\ α}\right)=\\ =\left({ch\ iα-sh\ iα}\right)\left({ch\ iα+sh\ iα}\right)=\\ =ch^2\ φ-sh^2\ φ\]

\[Tan^k\ α=Cot^{-k}\ α=\frac{Sin^k\ α}{Cos^k\ α}\]


\[Sin\left({2α}\right)=2·Sin\ α·Cos\ α\]

\[Cos\left({2α}\right)=Cos^2\ α-Sin^2\ α=\\ =2·Cos^2\ α-1=1-2·Sin^2\ α\]

\[Tan\left({2α}\right)=\frac{2·Tan\ α}{1-Tan^2\ α}\]

\[Cot\left({2α}\right)=\frac{Cot^2\ α-1}{2·Cot\ α}\]


\[1-Cos\ α=2·Sin^2\frac α2\]

\[1+Cos\ α=2·Cos^2\frac α2\]

\[1-Sin\ α=\left({Cos\frac α2-Sin\frac α2}\right)^2\]

\[1+Sin\ α=\left({Cos\frac α2+Sin\frac α2}\right)^2\]


\[Sin\frac α2=±\sqrt{\frac{1-Cos\ α}2}\]

\[Cos\frac α2=±\sqrt{\frac{1+Cos\ α}2}\]

\[Tan\frac α2=±\sqrt{\frac{1-Cos\ α}{1+Cos\ α}}=\frac{1-Cos\ α}{Sin\ α}=\frac{Sin\ α}{1+Cos\ α}\]


\[1+Tan^2\ α=\frac1{Cos^2\ α}\]

\[1+Cot^2\ α=\frac1{Sin^2\ α}\]

\[Sec\ α=\frac1{Cos\ α}=±\sqrt{1+Tan^2\ α}\]

\[Cosec\ α=\frac1{Sin\ α}=±\sqrt{1+Cot^2\ α}\]


.\(Sin\ α\)\(Cos\ α\)\(Tan\ α\)\(Cot\ α\)
\(Sin\ α\).\(\left({1-Cos^2\ α}\right)^{\frac12}\)\(\left({Tan^{-2}\ α+1}\right)^{-\frac12}\)\(\left({1+Cot^2\ α}\right)^{-\frac12}\).
\(Cos\ α\)\(\left({1-Sin^2\ α}\right)^{\frac12}\).\(\left({1+Tan^2\ α}\right)^{-\frac12}\)\(\left({1+Cot^{-2}\ α}\right)^{-\frac12}\)
\(Tan\ α\)\(\left({Sin^{-2}\ α-1}\right)^{-\frac12}\)\(\left({Cos^2\ α-1}\right)^{\frac12}\).\(Cot^{-1}\ α\)
\(Cot\ α\)\(\left({Sin^{-2}\ α-1}\right)^{\frac12}\)\(\left({Cos^2\ α-1}\right)^{-\frac12}\)\(Tan^{-1}\ α\).


\(Cos\ φ\)
\(0\)\(-1\)\(0\)\(1\)\(Cos\ φ\)\(\frac{\sqrt{2+\sqrt{2+\sqrt{3}}}}2\)\(\frac{\sqrt{2+\sqrt{3}}}2\)\(\frac{\sqrt{2+\sqrt{2}}}2\)\(\frac{\sqrt{3}}2\)\(\frac{\sqrt{2+\sqrt{2-\sqrt{3}}}}2\)\(\frac{\sqrt{2}}2\)\(\frac{\sqrt{2-\sqrt{2-\sqrt{3}}}}2\)\(\frac 12\)\(\frac{\sqrt{2-\sqrt{2}}}2\)\(\frac{\sqrt{2-\sqrt{3}}}2\)\(\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}2\)\(Cos\ φ\)
\(Sin\ φ\)\(-1\)\(0\)\(1\)\(0\)\(Sin\ φ\)\(\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}2\)\(\frac{\sqrt{2-\sqrt{3}}}2\)\(\frac{\sqrt{2-\sqrt{2}}}2\)\(\frac 12\)\(\frac{\sqrt{2-\sqrt{2-\sqrt{3}}}}2\)\(\frac{\sqrt{2}}2\)\(\frac{\sqrt{2+\sqrt{2-\sqrt{3}}}}2\)\(\frac{\sqrt{3}}2\)\(\frac{\sqrt{2+\sqrt{2}}}2\)\(\frac{\sqrt{2+\sqrt{3}}}2\)\(\frac{\sqrt{2+\sqrt{2+\sqrt{3}}}}2\)\(Sin\ φ\)
\(φ\)\(\frac{3π}2\)\(π\)\(\frac π2\)\(2kπ\)\(φ\)\(\frac{π}{24}\)\(\frac{π}{12}\)\(\frac π8\)\(\frac π6\)\(\frac{5π}{24}\)\(\frac π4\)\(\frac{7π}{24}\)\(\frac π3\)\(\frac{9π}{24}\)\(\frac{5π}{12}\)\(\frac{11π}{24}\)\(φ\)
\(Tan\ φ\)\(±∞\)\(0\)\(±∞\)\(0\)\(Tan\ φ\)\(\frac{\sqrt{2-\sqrt{3}}}{2+\sqrt{2+\sqrt{3}}}\)\(2-\sqrt{3}\)\(\frac1{\sqrt{2}+1}\)\(\frac1{\sqrt{3}}\)\(\frac{\sqrt{2+\sqrt{3}}}{2+\sqrt{2-\sqrt{3}}}\)\(1\)\(\frac{\sqrt{2+\sqrt{3}}}{2-\sqrt{2-\sqrt{3}}}\)\(\sqrt{3}\)\(\frac1{\sqrt{2}-1}\)\(2+\sqrt{3}\)\(\frac{\sqrt{2-\sqrt{3}}}{2-\sqrt{2+\sqrt{3}}}\)\(Tan\ φ\)
\(Cot\ φ\)\(0\)\(±∞\)\(0\)\(±∞\)\(Cot\ φ\)\(\frac{\sqrt{2-\sqrt{3}}}{2-\sqrt{2+\sqrt{3}}}\)\(2+\sqrt{3}\)\(\frac1{\sqrt{2}-1}\)\(\sqrt{3}\)\(\frac{\sqrt{2+\sqrt{3}}}{2-\sqrt{2-\sqrt{3}}}\)\(1\)\(\frac{\sqrt{2+\sqrt{3}}}{2+\sqrt{2-\sqrt{3}}}\)\(\frac1{\sqrt{3}}\)\(\frac1{\sqrt{2}+1}\)\(2-\sqrt{3}\)\(\frac{\sqrt{2-\sqrt{3}}}{2+\sqrt{2+\sqrt{3}}}\)\(Cot\ φ\)


\[arcCos\ A=±i·ln\left({A±\sqrt{A^2-1}}\right)\]

\[arcSin\ A=\left({±}\right)\ i·ln\left({±\sqrt{1-A^2}+\left({∓}\right)\ i·A}\right)\]

\[\frac i2\ ln\ \frac{1-i·A}{1+i·A}=arcTan\ A=\frac i2\ ln\ \frac{i+A}{i-A}\]

 \[\frac i2\ ln\ \frac{i·A+1}{i·A-1}=arcCot\ A=\frac i2\ ln\ \frac{A-i}{A+i}\]


\[arcTan\ A+arcCot\ A=\frac π2±2kπ\]

\[±\left({arcTan\ A-arcCot\ A}\right)=\frac i2\ ln\ \left({i·\frac{i·A∓1}{i·A±1}}\right)^2=\frac i2\ ln\ \left({-i·\frac{1∓i·A}{1±i·A}}\right)^2=\frac i2\ ln\ \left({i·\frac{A±i}{A∓i}}\right)^2=etc. ...\]


[Eop]

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