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# Right or Wrong?

Are all of these formulas right? If right or wrong then can you prove it?

*$$\large\sin (A+B)=\sin A\dot\ \cos B +\cos A\dot\ \sin B$$

*$$\large\sin (A-B)=\sin A\dot\ \cos B -\cos A\dot\ \sin B$$

*$$\large\cos (A+B)=\cos A\dot\ \cos B-\sin A\dot\ \sin B$$

*$$\large\cos (A-B)=\cos A\dot\ \cos B+\sin A\dot\ \sin B$$

*$$\large\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\dot\ \tan B}$$

*$$\large\tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A\dot\ \tan B}$$

*$$\large\cot(A+B)=\dfrac{\cot A\dot\ \cot B-1}{\cot B-\cot A}$$

*$$\large\cot(A-B)=\dfrac{\cot A\dot\ \cot B+1}{\cot A+\cot B}$$

*$$\large2\sin A\dot\ \cos B=\sin(A+B)+\sin(A-B)$$

*$$\large2\cos A\dot\ \sin B=\sin(A+B)-\sin(A-B)$$

*$$\large2\cos A\dot\ \cos B=\cos(A+B)+\cos(A-B)$$

*$$\large2\sin A\dot\ \sin B=\cos(A-B)-\cos(A+B)$$

*$$\large\sin (A+B)\dot\ \sin(A-B)=\sin^2 A-\sin^2 B=\cos^2 B-\cos^2 A$$

*$$\large\cos (A+B)\dot\ \cos(A-B)=\cos^2 A-\sin^2 B=\cos^2 B-\sin^2 A$$

*$$\large\sin C+\sin D=2\sin \dfrac{C+D}{2}\dot\ \sin \dfrac{C-D}{2}$$

*$$\large\sin C-\sin D=2\cos \dfrac{C+D}{2}\dot\ \sin \dfrac{C-D}{2}$$

*$$\large\cos C+\cos D=2\cos \dfrac{C+D}{2}\dot\ \cos \dfrac{C-D}{2}$$

*$$\large\cos C-\cos D=2\sin \dfrac{C+D}{2}\dot\ \sin \dfrac{D-C}{2}$$

*$$\large1-\cos 2A=2\sin^2 A$$

*$$\large1+\cos 2A=2\cos^2 A$$

*$$\large\dfrac{1-\cos 2A}{1+\cos 2A}=\tan 2A$$

*$$\large\sin 2A=2\sin A\cos A=\dfrac{2\tan A}{1+\tan^2 A}$$

*$$\large\cos 2A=(\cos^2 A-\sin^2 A)=(1-2\sin^2 A)=(2\cos^2 A-1)=(\dfrac{1-\tan^2 A}{1+\tan^2 A})$$

*$$\large\sin 2A=\dfrac{2\tan A}{1+tan^2 A}$$

*$$\large\cos 2A=\dfrac{1-\tan^2 A}{1+\tan^2 A}$$

*$$\large\sin 3A=3\sin A-4\sin^3 A$$

*$$\large\cos 3A=4\cos^3 A-3\cos A$$

*$$\large\tan 3A=\dfrac{3\tan A-\tan^3 A}{1-3\tan^2 A}$$

*$$\large\cos^3 A=\dfrac{3\cos A+\cos 3A}{4}$$

*$$\large\sin^3 A=\dfrac{3\sin A-\sin 3A}{4}$$

*$$\large\sin^{-1} x+\cos^{-1} x=\dfrac{\pi}{2}$$

*$$\large\tan^{-1} x+\cot^{-1} x=\dfrac{\pi}{2}$$

*$$\large\sec^{-1} x+\text{cosec}^{-1} x=\dfrac{\pi}{2}$$

*$$\large\tan^{-1} x+\tan^{-1} y=\tan^{-1}\dfrac{x+y}{1-xy}$$

*$$\large\tan^{-1} x-\tan^{-1} y=\tan^{-1}\dfrac{x-y}{1+xy}$$

*$$\large\tan^{-1} x+\tan^{-1} y+\tan^{-1} z= \tan^{-1}\dfrac{x+y+z-xyz}{1-yz-zx-xy}$$

*$$\large\sin^{-1} x+\sin^{-1} y=\sin^{-1} ({x\sqrt{1-y^2}+y\sqrt{1-x^2}})$$

*$$\large\sin^{-1} x-\sin^{-1} y=\sin^{-1}(x\sqrt{1-y^2}-y\sqrt{1-x^2})$$

*$$\large\cos^{-1} x+\cos^{-1} y=\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)})$$

*$$\large\cos^{-1} x-\cos^{-1} y=\cos^{-1}(xy+\sqrt{(1-x^2)(1-y^2)}$$

*$$\large2\tan^{-1} x=\sin^{-1}\dfrac{2x}{1+x^2}=\cos^{-1}\dfrac{1-x^2}{1+x^2}=\tan^{-1}\dfrac{2x}{1-x^2}$$

*$$\large\tan^{-1} x=\cot^{-1} \dfrac{1}{x}$$

*$$\large3\sin^{-1} x=\sin^{-1}(3x-4x^3)$$

*$$\large3\cos^{-1} x=\cos^{-1}(4x^3-3x)$$

*$$\large3\tan^{-1} x=\tan^{-1}\dfrac{3x-x^3}{1-3x^2}$$

Note by Sakib Nazmus
4 months, 1 week ago

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How can you be bothered to type this all up? I would be like zzzzzzzzzzzzz......

- 4 months ago