The sum formula and difference formula for \(\sin, \cos, \tan\):
\[ \begin{align} &\cos(\alpha+\beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta \\ &\cos(\alpha-\beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta \\ &\sin(\alpha+\beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta \\ &\sin(\alpha-\beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta \\ &\tan(\alpha+\beta) = \dfrac{1+\tan\alpha\tan\beta}{\tan\alpha\tan\beta} \\ &\tan(\alpha-\beta) = \dfrac{1-\tan\alpha\tan\beta}{\tan\alpha\tan\beta} \end{align} \]
The difference formula for tangent inferred that, for the two line with slope \(k_1, k_2\), the intersection angle can be computed by
\[ \tan\theta = \left| \frac{k_2-k_1}{1+k_1k_2} \right| \]
Specially, let \(\alpha = \beta\) in the sum formula we got the double angle formual \[ \begin{align} &\sin2\alpha = 2\sin\alpha\cos\alpha \\ &\cos2\alpha = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha \\ &\tan2\alpha = \dfrac{2\tan\alpha}{1-\tan^2\alpha} \end{align} \]
With identical transformation this can be applied to remove exponent from the trigonometric function: \[ \begin{align} &\sin^2 \alpha = \frac{1}{2}(\cos2\alpha-1) \\ &\cos^2 \alpha = \frac{1}{2}(\cos2\alpha+1) \\ &\tan^2 \alpha = 1 - \frac{2\tan\alpha}{\tan 2\alpha} \end{align} \] Furthermore we can get triple angle formula \[ \begin{align} \sin 3\alpha &= 3\sin\alpha - 4\sin^2\alpha \\ \cos 3\alpha &= -3\cos\alpha + 4\sin^2\alpha \\ \tan 3\alpha &= \frac{3\tan\alpha - \tan^3\alpha}{1-3\tan^2\alpha} = \tan\alpha \tan\left( \frac{\pi}{3} + \alpha \right) \tan\left( \frac{\pi}{3} - \alpha \right) \end{align} \]
\[ \begin{align} & \sin\alpha \cos\beta = \frac{\sin(\alpha+\beta) + \sin(\alpha-\beta)}{2} \\ & \cos\alpha \cos\beta = \frac{\cos(\alpha+\beta) + \cos(\alpha-\beta)}{2} \\ & \sin\alpha \sin\beta = \frac{-\cos(\alpha+\beta) + \cos(\alpha-\beta)}{2} \\ \end{align} \]
\[ \begin{align} & \sin\alpha + \sin\beta = 2\sin{\frac{\alpha+\beta}{2}} \cos{\frac{\alpha-\beta}{2}} \\ & \sin\alpha - \sin\beta = 2\cos{\frac{\alpha+\beta}{2}} \sin{\frac{\alpha-\beta}{2}} \\ & \cos\alpha + \cos\beta = 2\cos{\frac{\alpha+\beta}{2}} \cos{\frac{\alpha-\beta}{2}} \\ & \cos\alpha - \cos\beta = -2\sin{\frac{\alpha+\beta}{2}} \sin{\frac{\alpha-\beta}{2}} \\ \end{align} \]
\[ \begin{align} & \sin(A+B) = \sin C,~~~~ \cos(A+B) = -\cos C \\ & \sin\frac{A+B}{2} = \cos\frac{C}{2}, ~~~~ \cos\frac{A+B}{2} = \sin\frac{C}{2} \end{align} \]
\[ \begin{align} & \sin A + \sin B + \sin C = 4\cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2} \\ & \cos A + \cos B + \cos C = 1 + 4\sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} \\ & \tan A + \tan B + \tan C = \tan A \tan B \tan C \end{align} \]
Proof . \[ \begin{align} \sin A + \sin B + \sin C &= 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} + \sin(A+B) \\ &= 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} + 2\sin\frac{A+B}{2}\cos\frac{A+B}{2} \\ &= 4\sin\frac{A+B}{2} \cos\frac{A}{2}\cos\frac{B}{2} \\ &= 4\cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2} \end{align} \] The property of sum of cosine can be proved similarly.
For tangent, we have \[ \tan C = \tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B} \] multiplying \(1-\tan A\tan B\) on both sides: \[ \tan C - \tan A\tan B\tan C = \tan A+\tan B \] which gives the formula above.
We can introduce amounts of properties from the equation of tangent, i.e. with the help of AM-GM inequality we have \[ \tan A \tan B \tan C \ge 3\sqrt3 \] Both sides are equal if and only if its equilateral triangle.