\( \cos^2 x+\sin^2 x=1 \)
\( \tan x=\displaystyle\frac{\sin x}{\cos x} \)
\( 1+\tan^2 x=\displaystyle\frac{1}{\cos^2 x} \)
\( \cos(a+b)=\cos a \cos b - \sin a \sin b \)
\( \sin(a+b)=\cos a \sin b + \cos b \sin a \)
\( \tan(a+b)=\displaystyle\frac{\tan a + \tan b}{1-\tan a \tan b} \)
\( \cos(a-b)=\cos a \cos b + \sin a \sin b \)
\( \sin(a-b)=\cos b \sin a - \cos a \sin b \)
\( \tan(a-b)=\displaystyle\frac{\tan a - \tan b}{1+\tan a \tan b} \)
\( \cos(2a)=\cos^2 a - \sin^2 a = 2\cos^2 a -1 = 1-2\sin^2 a \)
\( \sin(2a)=2\sin a \cos a \)
\( \tan(2a)=\displaystyle\frac{2\tan a}{1-\tan^2 a} \)
\( \cos^2a= \displaystyle\frac{1+\cos(2a)}{2} \)
\( \sin^2a= \displaystyle\frac{1-\cos(2a)}{2} \)
\( \cos(a)\cos(b)=\frac{1}{2}\left(\cos(a+b)+\cos(a-b)\right) \)
\( \sin(a)\sin(b)=\frac{1}{2}\left(\cos(a-b)-\cos(a+b)\right) \)
\( \sin(a)\cos(b)=\frac{1}{2}\left(\sin(a-b)+\sin(a+b)\right) \)
\( \cos(p)+\cos(q)= 2\cos\left(\displaystyle\frac{p+q}{2}\right)\cos\left(\displaystyle\frac{p-q}{2}\right) \)
\( \cos(p)-\cos(q)=-2\sin\left(\displaystyle\frac{p+q}{2}\right)\sin\left(\displaystyle\frac{p-q}{2}\right) \)
\( \sin(p)+\sin(q)= 2\sin\left(\displaystyle\frac{p+q}{2}\right)\cos\left(\displaystyle\frac{p-q}{2}\right) \)
\( \sin(p)-\sin(q)= 2\sin\left(\displaystyle\frac{p-q}{2}\right)\cos\left(\displaystyle\frac{p+q}{2}\right) \)