Trigonometri-identiteter

[Karl_Erik]

Vis de følgende trigonometriske identitetene. (I de tre siste er det gitt at [tex]A+B+C=180^{\circ}[/tex].)

(i) [tex]sin X + sin Y = 2 sin {\frac {X+Y} 2} cos {\frac {X-Y} 2}[/tex]

(ii) [tex]cos X + cos Y = 2 cos {\frac {X+Y} 2} cos {\frac {X-Y} 2}[/tex]

(iii) [tex]cos A + cos B + cos C = 4 sin {\frac A 2} sin {\frac B 2} sin {\frac C 2} + 1[/tex]

(iiii) [tex]sin A + sin B + sin C = 4 cos {\frac A 2} cos {\frac B 2} cos {\frac C 2}[/tex]

(iiiii) [tex] tan A + tan B + tan C = tan A tan B tan C[/tex]

[Janhaa]

(iiii)
[tex]=\sin(A)+\sin(B)+\sin(180^0-A-B)\\=\sin(A)+\sin(B)+\sin(A+B)\\=\sin(A)+\sin(B)+\sin(A)\cos(B)+\cos(A)\sin(B)\\=\sin(A)(1+\cos(B))\,+\,\sin(B)(1+\cos(A))\\=2\sin({A\over 2})\cos({A\over 2})2\cos^2({B\over 2})+2\sin({B\over 2})\cos({B\over 2})2\cos^2({A\over 2})\\=4\cos({A\over 2})\cos({B\over 2})\left(\sin({A\over 2})\cos({B\over 2})\,+\,\sin({B\over 2})\cos({A\over 2})\right)\\=4\cos({A\over 2})\cos({B\over 2})\sin(\frac{A+B}{2})\\=4\cos({A\over 2})\cos({B\over 2})\sin(\frac{90^o-C}{2})\\=4\cos({A\over 2})\cos({B\over 2})\cos({C\over 2})[/tex]

[Janhaa]

(iiiii)
[tex]A+B+C=180^o[/tex]

[tex]\tan(A+B)=\tan(180^o-C)\\\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}=-\tan(C)\\\tan(A)+\tan(B)=(-\tan(C))(1-\tan(A)\tan(B))\\tan(A)+\tan(B)=-\tan(C)+\tan(A)\tan(B)\tan(C)\\\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)[/tex]