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

wahh maaf tulisan ancur. nulis di krl susah. nanti dirumah ketik ulang deh
[tex]\cos{A}-\cos{B}+\cos{C}-\cos{D}=-2\sin{\left(\frac{A+B}{2}\right)}\sin{\left(\frac{A-B}{2}\right)}-2\sin{\left(\frac{C+D}{2}\right)}\sin{\left(\frac{C-D}{2}\right)}[/tex]
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[tex]\sin{\left(\frac{C+D}{2}\right)}=\sin{\left(\frac{2\pi-(A+B)}{2}\right)}=\sin{\left(\pi-\frac{A+B}{2}\right)}=\sin{\left(\frac{A+B}{2}\right)}[/tex]
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[tex]-2\sin{\left(\frac{A+B}{2}\right)}\sin{\left(\frac{A-B}{2}\right)}-2\sin{\left(\frac{C+D}{2}\right)}\sin{\left(\frac{C-D}{2}\right)}=-2\sin{\left(\frac{A+B}{2}\right)}\sin{\left(\frac{A-B}{2}\right)}-2\sin{\left(\frac{A+B}{2}\right)}\sin{\left(\frac{C-D}{2}\right)}[/tex]
[tex]=-4\sin{\left(\frac{A+B}{2}\right)\times\left[\sin{\left(\frac{A-B}{2}\right)}+\sin{\left(\frac{C-D}{2}\right)}\right][/tex]
[tex]=-2\sin{\left(\frac{A+B}{2}\right)\times\left[2\sin{\left(\frac{A-B+C-D}{4}\right)}\cos{\left(\frac{A-B-C+D}{4}\right)}\right][/tex]
[tex]=-4\sin{\left(\frac{A+B}{2}\right)\sin{\left(\frac{A-B+C-D}{4}\right)}\cos{\left(\frac{A-B-C+D}{4}\right)}[/tex]
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[tex]\sin{\left(\frac{A-B+C-D}{4}\right)}=\sin{\left(\frac{(A+C)-(B+D)}{4}\right)}=\sin{\left(\frac{(A+C)-(2\pi-(A+C))}{4}\right)}=\sin{\left(\frac{-2pi+2(A+C)}{4}\right)}=-\cos{\left(\frac{A+C}{2}\right)}[/tex]

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