Jika A + B + C + D = 2π, tunjukkan bahwa: [tex]\displaystyle \sin A-\sin B+\sin C-\sin D=\\ -4\cos \left ( \frac{A+B}{2} \right )\sin \left ( \frac{A+C}{2} \rig

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

[tex]\cos{(\frac{C+D}{2})}=\cos{(\frac{2\pi-(A+B)}{2})}[/tex]
[tex]=\cos{(\pi-\frac{A+B}{2})}[/tex]
[tex]=-\cos{(\frac{A+B}{2})}[/tex]

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

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

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

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