Right or Wrong?

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

*sin(A+B)=sinA ˙cosB+cosA ˙sinB\large\sin (A+B)=\sin A\dot\ \cos B +\cos A\dot\ \sin B

*sin(AB)=sinA ˙cosBcosA ˙sinB\large\sin (A-B)=\sin A\dot\ \cos B -\cos A\dot\ \sin B

*cos(A+B)=cosA ˙cosBsinA ˙sinB\large\cos (A+B)=\cos A\dot\ \cos B-\sin A\dot\ \sin B

*cos(AB)=cosA ˙cosB+sinA ˙sinB\large\cos (A-B)=\cos A\dot\ \cos B+\sin A\dot\ \sin B

*tan(A+B)=tanA+tanB1tanA ˙tanB\large\tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\dot\ \tan B}

*tan(AB)=tanAtanB1+tanA ˙tanB\large\tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A\dot\ \tan B}

*cot(A+B)=cotA ˙cotB1cotBcotA\large\cot(A+B)=\dfrac{\cot A\dot\ \cot B-1}{\cot B-\cot A}

*cot(AB)=cotA ˙cotB+1cotA+cotB\large\cot(A-B)=\dfrac{\cot A\dot\ \cot B+1}{\cot A+\cot B}

*2sinA ˙cosB=sin(A+B)+sin(AB)\large2\sin A\dot\ \cos B=\sin(A+B)+\sin(A-B)

*2cosA ˙sinB=sin(A+B)sin(AB)\large2\cos A\dot\ \sin B=\sin(A+B)-\sin(A-B)

*2cosA ˙cosB=cos(A+B)+cos(AB)\large2\cos A\dot\ \cos B=\cos(A+B)+\cos(A-B)

*2sinA ˙sinB=cos(AB)cos(A+B)\large2\sin A\dot\ \sin B=\cos(A-B)-\cos(A+B)

*sin(A+B) ˙sin(AB)=sin2Asin2B=cos2Bcos2A\large\sin (A+B)\dot\ \sin(A-B)=\sin^2 A-\sin^2 B=\cos^2 B-\cos^2 A

*cos(A+B) ˙cos(AB)=cos2Asin2B=cos2Bsin2A\large\cos (A+B)\dot\ \cos(A-B)=\cos^2 A-\sin^2 B=\cos^2 B-\sin^2 A

*sinC+sinD=2sinC+D2 ˙sinCD2\large\sin C+\sin D=2\sin \dfrac{C+D}{2}\dot\ \sin \dfrac{C-D}{2}

*sinCsinD=2cosC+D2 ˙sinCD2\large\sin C-\sin D=2\cos \dfrac{C+D}{2}\dot\ \sin \dfrac{C-D}{2}

*cosC+cosD=2cosC+D2 ˙cosCD2\large\cos C+\cos D=2\cos \dfrac{C+D}{2}\dot\ \cos \dfrac{C-D}{2}

*cosCcosD=2sinC+D2 ˙sinDC2\large\cos C-\cos D=2\sin \dfrac{C+D}{2}\dot\ \sin \dfrac{D-C}{2}

*1cos2A=2sin2A\large1-\cos 2A=2\sin^2 A

*1+cos2A=2cos2A\large1+\cos 2A=2\cos^2 A

*1cos2A1+cos2A=tan2A\large\dfrac{1-\cos 2A}{1+\cos 2A}=\tan 2A

*sin2A=2sinAcosA=2tanA1+tan2A\large\sin 2A=2\sin A\cos A=\dfrac{2\tan A}{1+\tan^2 A}

*cos2A=(cos2Asin2A)=(12sin2A)=(2cos2A1)=(1tan2A1+tan2A)\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})

*sin2A=2tanA1+tan2A\large\sin 2A=\dfrac{2\tan A}{1+tan^2 A}

*cos2A=1tan2A1+tan2A\large\cos 2A=\dfrac{1-\tan^2 A}{1+\tan^2 A}

*sin3A=3sinA4sin3A\large\sin 3A=3\sin A-4\sin^3 A

*cos3A=4cos3A3cosA\large\cos 3A=4\cos^3 A-3\cos A

*tan3A=3tanAtan3A13tan2A\large\tan 3A=\dfrac{3\tan A-\tan^3 A}{1-3\tan^2 A}

*cos3A=3cosA+cos3A4\large\cos^3 A=\dfrac{3\cos A+\cos 3A}{4}

*sin3A=3sinAsin3A4\large\sin^3 A=\dfrac{3\sin A-\sin 3A}{4}

*sin1x+cos1x=π2\large\sin^{-1} x+\cos^{-1} x=\dfrac{\pi}{2}

*tan1x+cot1x=π2\large\tan^{-1} x+\cot^{-1} x=\dfrac{\pi}{2}

*sec1x+cosec1x=π2\large\sec^{-1} x+\text{cosec}^{-1} x=\dfrac{\pi}{2}

*tan1x+tan1y=tan1x+y1xy\large\tan^{-1} x+\tan^{-1} y=\tan^{-1}\dfrac{x+y}{1-xy}

*tan1xtan1y=tan1xy1+xy\large\tan^{-1} x-\tan^{-1} y=\tan^{-1}\dfrac{x-y}{1+xy}

*tan1x+tan1y+tan1z=tan1x+y+zxyz1yzzxxy\large\tan^{-1} x+\tan^{-1} y+\tan^{-1} z= \tan^{-1}\dfrac{x+y+z-xyz}{1-yz-zx-xy}

*sin1x+sin1y=sin1(x1y2+y1x2)\large\sin^{-1} x+\sin^{-1} y=\sin^{-1} ({x\sqrt{1-y^2}+y\sqrt{1-x^2}})

*sin1xsin1y=sin1(x1y2y1x2)\large\sin^{-1} x-\sin^{-1} y=\sin^{-1}(x\sqrt{1-y^2}-y\sqrt{1-x^2})

*cos1x+cos1y=cos1(xy(1x2)(1y2))\large\cos^{-1} x+\cos^{-1} y=\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)})

*cos1xcos1y=cos1(xy+(1x2)(1y2)\large\cos^{-1} x-\cos^{-1} y=\cos^{-1}(xy+\sqrt{(1-x^2)(1-y^2)}

*2tan1x=sin12x1+x2=cos11x21+x2=tan12x1x2\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}

*tan1x=cot11x\large\tan^{-1} x=\cot^{-1} \dfrac{1}{x}

*3sin1x=sin1(3x4x3)\large3\sin^{-1} x=\sin^{-1}(3x-4x^3)

*3cos1x=cos1(4x33x)\large3\cos^{-1} x=\cos^{-1}(4x^3-3x)

*3tan1x=tan13xx313x2\large3\tan^{-1} x=\tan^{-1}\dfrac{3x-x^3}{1-3x^2}

Note by Nazmus Sakib
3 years, 9 months ago

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

Annie Li - 3 years, 9 months ago

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