Pertanyaan

Jika A + B + C = 2 S , tunjukkan bahwa. cos 2 S + cos 2 ( S − A ) + cos 2 ( S − B ) + cos 2 ( S − C ) = 2 + 2 cos A cos B cos C

Jika $A + B + C = 2 S$ , tunjukkan bahwa.

$cos^{2} S + cos^{2} (S - A) + cos^{2} (S - B) + cos^{2} (S - C) = 2 + 2 cos A cos B cos C$

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Habis dalam

Klaim

I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

Ingat bahwa : Rumus jumlah dan selisih Trigonometri yaitu cos A + cos B = 2 cos 2 1 ( A + B ) cos 2 1 ( A − B ) Sudut rangkap pada cosinus cos 2 A 1 + cos 2 A 2 1 + c o s 2 A = = = 2 cos 2 A − 1 2 cos 2 A cos 2 A Dari soal diketahui A + B + C A cos A = = = 2 S 2 S − ( B + C ) cos ( 2 S − ( B + C ) ) Sehingga cos 2 S + cos 2 ( S − A ) + cos 2 ( S − B ) + cos 2 ( S − C ) ⇔ 2 1 + cos 2 S + 2 1 + cos 2 ( S − A ) + 2 1 + cos 2 ( S − B ) + 2 1 + cos 2 ( S − C ) ⇔ 2 1 [ 4 + cos 2 S + cos ( 2 S − 2 A ) + cos ( 2 S − 2 B ) + cos ( 2 S − 2 C ) ] ⇔ 2 + 2 1 [ 2 cos 2 1 ( 2 S + 2 S − 2 A ) cos 2 1 ( 2 S − 2 S + 2 A ) + 2 cos 2 1 ( 2 S − 2 B + 2 S − 2 C ) cos 2 1 ( 2 S − 2 B − 2 S + 2 C ) ] ⇔ 2 + cos ( 2 4 S − 2 A ) cos 2 2 A + cos ( 2 4 S − 2 ( B + C ) ) cos ( 2 2 ( C − B ) ) ⇔ 2 + cos ( 2 S − A ) cos A + cos ( 2 S − ( B + C ) ) cos ( C − B ) ⇔ 2 + cos ( 2 S − A ) cos A + cos A cos ( C − B ) ⇔ 2 + cos A ( cos ( 2 S − A ) + cos ( C − B ) ) ⇔ 2 + cos A ( cos ( B + C ) + cos ( C − B ) ) ⇔ 2 + cos A ( 2 cos ( 2 B + C + C − B ) cos ( 2 B + C − C + B ) ) ⇔ 2 + cos A cos 2 2 C cos 2 2 B ⇔ 2 + 2 cos A cos B cos C Dengan demikian terbuktibahwa cos 2 S + cos 2 ( S − A ) + cos 2 ( S − B ) + cos 2 ( S − C ) = 2 + 2 cos A cos B cos C

Ingat bahwa :

Rumus jumlah dan selisih Trigonometri yaitu

$cos A + cos B = 2 cos \frac{1}{2} (A + B) cos \frac{1}{2} (A - B)$

Sudut rangkap pada cosinus

$cos 2 A 1 + cos 2 A \frac{1 + c o s 2 A}{2} = = = 2 cos^{2} A - 1 2 cos^{2} A cos^{2} A$

Dari soal diketahui

$A + B + C A cos A = = = 2 S 2 S - (B + C) cos (2 S - (B + C))$

Sehingga

$cos^{2} S + cos^{2} (S - A) + cos^{2} (S - B) + cos^{2} (S - C) \Leftrightarrow \frac{1 + cos 2 S}{2} + \frac{1 + cos 2 ( S - A )}{2} + \frac{1 + cos 2 ( S - B )}{2} + \frac{1 + cos 2 ( S - C )}{2} \Leftrightarrow \frac{1}{2} [4 + cos 2 S + cos (2 S - 2 A) + cos (2 S - 2 B) + cos (2 S - 2 C)] \Leftrightarrow 2 + \frac{1}{2} [2 cos \frac{1}{2} (2 S + 2 S - 2 A) cos \frac{1}{2} (2 S - 2 S + 2 A) + 2 cos \frac{1}{2} (2 S - 2 B + 2 S - 2 C) cos \frac{1}{2} (2 S - 2 B - 2 S + 2 C)] \Leftrightarrow 2 + cos (\frac{4 S - 2 A}{2}) cos \frac{2 A}{2} + cos (\frac{4 S - 2 ( B + C )}{2}) cos (\frac{2 ( C - B )}{2}) \Leftrightarrow 2 + cos (2 S - A) cos A + cos (2 S - (B + C)) cos (C - B) \Leftrightarrow 2 + cos (2 S - A) cos A + cos A cos (C - B) \Leftrightarrow 2 + cos A (cos (2 S - A) + cos (C - B)) \Leftrightarrow 2 + cos A (cos (B + C) + cos (C - B)) \Leftrightarrow 2 + cos A (2 cos (\frac{B + C + C - B}{2}) cos (\frac{B + C - C + B}{2})) \Leftrightarrow 2 + cos A cos \frac{2 C}{2} cos \frac{2 B}{2} \Leftrightarrow 2 + 2 cos A cos B cos C$