Question

sum and difference formula and double angle formula for inverse trigonomertry.

Answer 1

$$\begin{gathered} \mathbf{SUM}\mathbf{}\mathbf{FORMULAS}\mathbf{}: \\ \sin \left(\mathrm{A}+\mathrm{B}\right)=\sin \mathrm{A}.\cos \mathrm{B}+\cos \mathrm{A}.\sin \mathrm{B} \\ \cos \left(\mathrm{A}+\mathrm{B}\right)=\cos \mathrm{A}.\cos \mathrm{B}-\sin \mathrm{A}.\sin \mathrm{B} \\ \tan \left(\mathrm{A}+\mathrm{B}\right)=\frac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A}.\tan \mathrm{B}} \\ \mathbf{DIFFERENCE}\mathbf{}\mathbf{FORMULAS}: \\ \sin \left(\mathrm{A}-\mathrm{B}\right)=\sin \mathrm{A}.\cos \mathrm{B}-\cos \mathrm{A}.\sin \mathrm{B} \\ \cos \left(\mathrm{A}-\mathrm{B}\right)=\cos \mathrm{A}.\cos \mathrm{B}+\sin \mathrm{A}.\sin \mathrm{B} \\ \tan \left(\mathrm{A}-\mathrm{B}\right)=\frac{\tan \mathrm{A}-\tan \mathrm{B}}{1+\tan \mathrm{A}.\tan \mathrm{B}} \\ \mathbf{DOUBLE}\mathbf{}\mathbf{ANGLE}\mathbf{}\mathbf{FORMULAS}: \\ \sin 2\mathrm{A}=2\sin \mathrm{A}.\cos \mathrm{A}=\frac{2\tan \mathrm{A}}{1+{\tan }^2\mathrm{A}} \\ \cos 2\mathrm{A}={\cos }^2\mathrm{A}-{\sin }^2\mathrm{A}=1-2{\sin }^2\mathrm{A}=2{\cos }^2\mathrm{A}-1=\frac{1-{\tan }^2\mathrm{A}}{1+{\tan }^2\mathrm{A}} \\ \tan 2\mathrm{A}=\frac{2\tan \mathrm{A}}{1-{\tan }^2\mathrm{A}} \end{gathered}$$