Pembahasan trigonometri (6)


Untuk soal nomor 9
jika \alpha+\beta+\gamma=90^{o}, maka \tan\alpha\tan\beta+\tan\alpha\tan\gamma+\tan\beta\tan\gamma=1
bukti
karena \alpha+\beta+\gamma=90^{o}, maka kita peroleh :
\tan(\alpha+\beta)=\tan(90^{o}-\gamma)
\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{1}{\tan\gamma}
(\tan\alpha+\tan\beta)\tan\gamma=1-\tan\alpha\tan\beta
\tan\alpha\tan\gamma+\tan\beta\tan\gamma=1-\tan\alpha\tan\beta
\tan\alpha\tan\gamma+\tan\beta\tan\gamma+\tan\alpha\tan\beta=1

Terbukti

untuk soal nomor 10
jika \sin\alpha\cos^{2}\beta=\sin\beta\cos^{2}\alpha buktikan \Delta ABC samakaki

bukti

\sin\alpha\cos^{2}\beta=\sin\beta\cos^{2}\alpha
\sin\alpha(1-\sin^{2}\beta)=\sin\beta(1-\sin^{2}\alpha)
\sin\alpha-\sin\alpha\sin^{2}\beta=\sin\beta-\sin\beta\sin^{2}\alpha)
\sin\alpha+\sin\beta\sin^{2}\alpha=\sin\beta+\sin\alpha\sin^{2}\beta
\sin\alpha(1+\sin\beta\sin\alpha)=\sin\beta(1+\sin\alpha\sin\beta)
kedua ruas dibagi dengan (1+\sin\beta\sin\alpha), sehingga diperoleh
\sin\alpha=\sin\beta sehingga \alpha=\beta
Jadi \Delta ABC sama kaki
Terbukti.

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