Pembahasan trigonometri (4)


Untuk soal nomor 6.
jika \alpha+\beta+\gamma=180^{o} , maka
\sin(\alpha+\beta)=\sin\gamma….. (1),
langkah 1
\beta+ \gamma=180^{o}-\alpha , sehingga
\cos(\beta+\gamma)=\cos(180^{o}-\alpha)=-\cos\alpha
\cos(\beta+\gamma)=-\cos\alpha
kedua ruas dikalikan dengan \cos\alpha, sehingga
\cos(\beta+\gamma).\cos\alpha=-\cos^{2}\alpha
(\cos\beta\cos\gamma -\sin\beta\sin\gamma)\cos\alpha=-\cos^{2}\alpha
\cos\alpha\cos\beta\cos\gamma-\cos\alpha\sin\beta\sin\gamma=-\cos^{2}\alpha
\cos^{2}\alpha -\cos\alpha\sin\beta\sin\gamma=-\cos\alpha\cos\beta\cos\gamma... (2)
langkah 1 diulang
\cos(\alpha+\gamma)=\cos(180^{o}-\beta)=-\cos\beta
\cos(\alpha+\gamma)=-\cos\beta
kedua ruas dikalikan dengan \cos\beta, sehingga
\cos(\alpha+\gamma)\cos\beta=-\cos^{2}\beta
(\cos\alpha\cos\gamma-\sin\alpha\sin\gamma)\cos\beta=-\cos^{2}\beta
\cos\alpha\cos\beta\cos\gamma-\sin\alpha\cos\beta\sin\gamma=-\cos^{2}\beta
\cos^{2}\beta-\sin\alpha\cos\beta\sin\gamma=-\cos\alpha\cos\beta\cos\gamma ... (3)
dari persamaan (2) dijumlah dengan persamaan (3)
\cos^{2}\alpha-\cos\alpha\sin\beta\sin\gamma+\cos^{2}\beta-\sin\alpha\cos\beta\sin\gamma=
-2\cos\alpha\cos\beta\cos\gamma
\cos^{2}\alpha+\cos^{2}\beta-\sin\gamma(\sin\alpha\cos\beta+\cos\alpha\sin\beta)=
-2\cos\alpha\cos\beta\cos\gamma
\cos^{2}\alpha+\cos^{2}\beta-\sin\gamma\sin(\alpha+\beta)=-2\cos\alpha\cos\beta\cos\gamma
subtitusi persamaan (1)
\cos^{2}\alpha+\cos^{2}\beta-\sin^{2}\gamma=-2\cos\alpha\cos\beta\cos\gamma
\cos^{2}\alpha+\cos^{2}\beta-(1-\cos^{2}\gamma)=-2\cos\alpha\cos\beta\cos\gamma
\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1-2\cos\alpha\cos\beta\cos\gamma
terbukti

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