# Trigonometry - Relations between trigonometric functions

1. $sin\alpha=\pm\sqrt{1-cos^2\alpha}=\pm\sqrt{{1\over2}(1-cos2\alpha)}$ =$2cos^2({\alpha \over 2}-{\pi \over 2})-1$=${2tan({\alpha \over 2})} \over 1+tan^2 {\alpha \over2}$
2. $cos\alpha=\pm\sqrt{1-sin^2\alpha}=\pm\sqrt{{1\over2}(1+cos2\alpha)}$$2cos^2{\alpha \over 2}-1$${1-tan^2{\alpha \over 2}} \over 1+tan^2 {\alpha \over2}$
3. $tan\alpha={sin\alpha \over cos\alpha}=\pm \sqrt{1-sin^2\alpha}$=$sin2\alpha \over1+cos2\alpha$=$1-cos2\alpha \over sin2\alpha$=$2tan{\alpha \over 2} \over1+ 2tan^2{\alpha \over 2}$=${2tan{\alpha \over2}} \over1+tan^2{\alpha \over 2}$=${\pm\sqrt{1-cos2\alpha \over 1+cos2\alpha}}$
4. $cot\alpha={cos\alpha \over sin\alpha}$=$\pm\sqrt{sec^2-1}$=$1+cos2\alpha \over sin2 \alpha$=$\pm{\sqrt{{1+cos2\alpha} \over {1-cos2\alpha}}}$=${1-tan^2{\alpha \over2}} \over2tan^2{\alpha \over 2}$
5. $sec \alpha= {1 \over cos \alpha}$=$\pm\sqrt{1+tan^2\alpha}$=${1+tan^2{\alpha \over2}} \over {1-tan^2{\alpha \over2}}$
6. $cosec \alpha= {1 \over sin \alpha}$=$\pm \sqrt {1+cot^2 \alpha }$=${1+tan^2{\alpha \over2}} \over2tan{\alpha \over 2}$