Help on trigonometry please!

$\sin^2 x + \cos^2x = 1\\ 1 + \tan^2x = \sec^2x\\ 1+\cot^2x =\csc^2x\\ \cos(-z) = \cos z\\ \sin(-z) = -\sin z\\ \tan(-z) = -\tan z\\ \sin 2x =2\sin x \cos x\\ \cos 2x = 2\cos^2x - 1 = 1 - 2\sin^2x \tan 2x = \dfrac{2\tan x}{1-\tan^2x}\\ \sin^2x = \dfrac{1}{2}(1-\cos 2x)\\ \cos^2x = \dfrac{1}{2}(1+\cos 2x)\\ \cos(x + y) = \cos x \cos y - \sin x \sin y\\ \sin(x + y) = \sin x \cos y + \cos x \sin y\\ \tan(x+ y) = \dfrac{\tan x +\tan y}{1-\tan x \tan y}\\ \sin x - \sin y = 2\cos(\dfrac{x+y}{2})\sin)(\dfrac{x-y}{2})\\ \cos x - \cos y = -2\sin(\dfrac{x+y}{2})\sin(\dfrac{x-y}{2})\\ \cos x\cos y = \dfrac{1}{2} (\cos(x + y)+\cos(x-y))\\ \sin x \sin y = \dfrac{1}{2}(\cos(x-y)-\cos(x+y))\\ \cos x \sin y = \dfrac{1}{2}(\sin(x+y)-\sin(x-y))$