Pembahasan trigonometri (3)


untuk soal nomor 3
\sin x +\sin3x +\sin5x -\sin9x =

2 \sin 2x.\cos (-x) +2 \cos 7x \sin(-2x) =

2 \sin 2x.\cos x  -2 \cos 7x \sin 2x =

2 \sin 2x(\cos x  -\cos 7x)=

2 \sin 2x(-2 \sin 4x \sin (-3x)=

4 \sin 2x \sin 4x \sin 3x

Untuk soal nomor 4

jika \alpha+\beta+\gamma = 90^{o}
\alpha+\beta = 90^{o}-\gamma
\sin(\alpha+\beta) = \sin(90-\gamma)^{o}
\sin(\alpha+\beta) = \cos\gamma ... (1)
analog dengan cara di atas kita dapatkan :
\sin(\alpha+\gamma) = \cos\beta... (2)
\sin(\beta+\gamma) = \cos\alpha ... (3)

dari yang diketahui kita dapatkan

\sin(\alpha+\beta+\gamma) = sin(90)^{o}
\sin((\alpha+\beta)+\gamma) = 1
\sin(\alpha+\beta)\cos\gamma + \cos(\alpha+\beta)\sin\gamma= 1
subtitusi (1)
\cos\gamma\cos\gamma + (\cos\alpha\cos\beta-\sin\alpha\sin\beta)\sin\gamma= 1
\cos^{2}\gamma + \cos\alpha\cos\beta\sin\gamma-\sin\alpha\sin\beta\sin\gamma= 1
\cos^{2}\gamma + \cos\alpha\cos\beta\sin\gamma = 1 +\sin\alpha\sin\beta\sin\gamma ... (4)
kita ulang langkah diatas

\sin(\alpha+\beta+\gamma) = sin(90)^{o}
\sin(\alpha+(\beta+\gamma)) = 1
\sin\alpha\cos(\beta+\gamma)+\cos\alpha\sin(\beta+\gamma)=1
subtitusi persamaan (3), sehingga
\sin\alpha\cos(\beta+\gamma)+\cos\alpha\cos\alpha=1
\sin\alpha(\cos\beta\cos\gamma-\sin\beta\sin\gamma)+\cos^{2}\alpha=1
\sin\alpha\cos\beta\cos\gamma-\sin\alpha\sin\beta\sin\gamma+\cos^{2}\alpha=1
\sin\alpha\cos\beta\cos\gamma+\cos^{2}\alpha=1+\sin\alpha\sin\beta\sin\gamma .... (5)
(4) + (5) kita dapatkan

\cos^{2}\alpha+\cos^{2}\gamma +\sin\alpha\cos\beta\cos\gamma+\cos\alpha\cos\beta\sin\gamma =
2+2\sin\alpha\sin\beta\sin\gamma
\cos^{2}\alpha+\cos^{2}\gamma+\cos\beta(\sin\gamma\cos\alpha+\cos\gamma\sin\alpha)=
2+2\sin\alpha\sin\beta\sin\gamma
\cos^{2}\alpha+\cos^{2}\gamma+\cos\beta\sin(\gamma+\alpha)=
2+2\sin\alpha\sin\beta\sin\gamma
subtitusi persamaan (2) sehingga
\cos^{2}\alpha+\cos^{2}\gamma+\cos\beta\cos\beta= 2 + 2\sin\alpha\sin\beta\sin\gamma
\cos^{2}\alpha+\cos^{2}\gamma+\cos^{2}\beta= 2(1+ \sin\alpha\sin\beta\sin\gamma)
Terbukti

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