Pertanyaan

9. Jika A + B + C = 18 0 ∘ , tunjukkan bahwa: c. cos 2 A + cos 2 B − cos 2 C = 1 − 4 sin A ⋅ sin B ⋅ cos C

9. Jika $A + B + C = 18 0^{\circ}$ , tunjukkan bahwa:

c. $cos 2 A + cos 2 B - cos 2 C = 1 - 4 sin A \cdot sin B \cdot cos C$ $\;$

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Jawaban terverifikasi

Jawaban

telah ditunjukan bahwa cos 2 A + cos 2 B − cos 2 C = 1 − 4 ⋅ sin A ⋅ sin B ⋅ cos C .

telah ditunjukan bahwa

cos 2 A + cos 2 B - cos 2 C = 1 - 4 \cdot sin A \cdot sin B \cdot cos C

. $\;$

Pembahasan

Ingat kembali: sudut berelasi: cos ( 180 − α ) = − cos α rumus cosinus untuk penjumlahan dua sudut: cos ( α + β ) = cos α ⋅ cos β − sin α ⋅ sin β r umus cosinus untuk sudut ganda: cos 2 α = 1 − 2 ⋅ sin 2 α = 2 cos 2 α − 1 identitas trigonometri: cos 2 α = 1 − sin 2 α , sin 2 α = 1 − cos 2 α , dan sin 2 α + cos 2 α = 1 Olehkarena itu, dengan menggunakan sudut berelasi dapatdiperoleh: cos C = = = = = cos ( 18 0 ∘ − ( A + B ) ) − cos ( A + B ) − ( cos A ⋅ cos B − sin A ⋅ sin B ) − cos A ⋅ cos B + sin A ⋅ sin B sin A ⋅ sin B − cos A ⋅ cos B sehingga diperoleh: cos 2 C = = = = = = = = = = ( cos C ) 2 ( sin A ⋅ sin B − cos A ⋅ cos B ) 2 ( sin A ⋅ sin B ) 2 + ( cos A ⋅ cos B ) 2 − 2 ⋅ sin A ⋅ sin B ⋅ cos A ⋅ cos B sin 2 A ⋅ sin 2 B + cos 2 A ⋅ cos 2 B − 2 ⋅ sin A ⋅ sin B ⋅ cos A ⋅ cos B sin 2 A ⋅ sin 2 B + ( 1 − sin 2 A ) ⋅ ( 1 − sin 2 B ) − 2 ⋅ sin A ⋅ sin B ⋅ cos A ⋅ cos B sin 2 A ⋅ sin 2 B + 1 − sin 2 A − sin 2 B + sin 2 A ⋅ sin 2 B − 2 ⋅ sin A ⋅ sin B ⋅ cos A ⋅ cos B 1 − sin 2 A − sin 2 B + sin 2 A ⋅ sin 2 B + sin 2 A ⋅ sin 2 B − 2 ⋅ sin A ⋅ sin B ⋅ cos A ⋅ cos B 1 − sin 2 A − sin 2 B + 2 ⋅ sin 2 A ⋅ sin 2 B − 2 ⋅ sin A ⋅ sin B ⋅ cos A ⋅ cos B 1 − sin 2 A − sin 2 B + 2 ⋅ sin A ⋅ sin B ( sin A ⋅ sin B − c os A ⋅ cos B ) 1 − sin 2 A − sin 2 B + 2 ⋅ sin A ⋅ sin B ⋅ cos C Selanjutnya, dengan menggunakan rumus sudut ganda dan kemudian menyubtitusi bentuk cos 2 C yang diperoleh di atas, maka diperoleh: = = = = = = = = = = = = cos 2 A + cos 2 B − cos 2 C 2 ⋅ cos 2 A − 1 + 2 ⋅ cos 2 B − 1 − ( 2 ⋅ cos 2 C − 1 ) 2 ⋅ cos 2 A − 1 + 2 ⋅ cos 2 B − 1 − 2 ⋅ cos 2 C + 1 2 ⋅ cos 2 A + 2 ⋅ cos 2 B − 2 ⋅ cos 2 C − 1 − 1 + 1 2 ⋅ cos 2 A + 2 ⋅ cos 2 B − 2 ⋅ cos 2 C − 1 2 ⋅ cos 2 A + 2 ⋅ cos 2 B − 2 ⋅ ( 1 − sin 2 A − sin 2 B + 2 ⋅ sin A ⋅ sin B ⋅ cos C ) − 1 2 ⋅ cos 2 A + 2 ⋅ cos 2 B − 2 + 2 ⋅ sin 2 A + 2 ⋅ sin 2 B − 4 ⋅ sin A ⋅ sin B ⋅ cos C − 1 − 2 − 1 + 2 ⋅ cos 2 A + 2 ⋅ sin 2 A + 2 ⋅ cos 2 B + 2 ⋅ sin 2 B − 4 ⋅ sin A ⋅ sin B ⋅ cos C − 3 + 2 ⋅ ( cos 2 A + sin 2 A ) + 2 ⋅ ( cos 2 B + sin 2 B ) − 4 ⋅ sin A ⋅ sin B ⋅ cos C − 3 + 2 ⋅ ( 1 ) + 2 ⋅ ( 1 ) − 4 ⋅ sin A ⋅ sin B ⋅ cos C − 3 + 2 + 2 − 4 ⋅ sin A ⋅ sin B ⋅ cos C − 3 + 4 − 4 ⋅ sin A ⋅ sin B ⋅ cos C 1 − 4 ⋅ sin A ⋅ sin B ⋅ cos C Dengan demikian, telah ditunjukan bahwa cos 2 A + cos 2 B − cos 2 C = 1 − 4 ⋅ sin A ⋅ sin B ⋅ cos C .

Ingat kembali:

sudut berelasi: $cos (180 - α) = - cos α$
rumus cosinus untuk penjumlahan dua sudut: $cos (α + β) = cos α \cdot cos β - sin α \cdot sin β$
rumus cosinus untuk sudut ganda: $cos 2 α = 1 - 2 \cdot sin^{2} α = 2 cos^{2} α - 1$
identitas trigonometri: $cos^{2} α = 1 - sin^{2} α$ , $sin^{2} α = 1 - cos^{2} α$ , dan $sin^{2} α + cos^{2} α = 1$

Oleh karena itu, dengan menggunakan sudut berelasi dapat diperoleh:

$cos C = = = = = cos (18 0^{\circ} - (A + B)) - cos (A + B) - (cos A \cdot cos B - sin A \cdot sin B) - cos A \cdot cos B + sin A \cdot sin B sin A \cdot sin B - cos A \cdot cos B$

sehingga diperoleh:

$cos^{2} C = = = = = = = = = = (cos C)^{2} (sin A \cdot sin B - cos A \cdot cos B)^{2} (sin A \cdot sin B)^{2} + (cos A \cdot cos B)^{2} - 2 \cdot sin A \cdot sin B \cdot cos A \cdot cos B sin^{2} A \cdot sin^{2} B + cos^{2} A \cdot cos^{2} B - 2 \cdot sin A \cdot sin B \cdot cos A \cdot cos B sin^{2} A \cdot sin^{2} B + (1 - sin^{2} A) \cdot (1 - sin^{2} B) - 2 \cdot sin A \cdot sin B \cdot cos A \cdot cos B sin^{2} A \cdot sin^{2} B + 1 - sin^{2} A - sin^{2} B + sin^{2} A \cdot sin^{2} B - 2 \cdot sin A \cdot sin B \cdot cos A \cdot cos B 1 - sin^{2} A - sin^{2} B + sin^{2} A \cdot sin^{2} B + sin^{2} A \cdot sin^{2} B - 2 \cdot sin A \cdot sin B \cdot cos A \cdot cos B 1 - sin^{2} A - sin^{2} B + 2 \cdot sin^{2} A \cdot sin^{2} B - 2 \cdot sin A \cdot sin B \cdot cos A \cdot cos B 1 - sin^{2} A - sin^{2} B + 2 \cdot sin A \cdot sin B (sin A \cdot sin B - c os A \cdot cos B) 1 - sin^{2} A - sin^{2} B + 2 \cdot sin A \cdot sin B \cdot cos C$

Selanjutnya, dengan menggunakan rumus sudut ganda dan kemudian menyubtitusi bentuk $cos^{2} C$ yang diperoleh di atas, maka diperoleh:

$= = = = = = = = = = = = cos 2 A + cos 2 B - cos 2 C 2 \cdot cos^{2} A - 1 + 2 \cdot cos^{2} B - 1 - (2 \cdot cos^{2} C - 1) 2 \cdot cos^{2} A - 1 + 2 \cdot cos^{2} B - 1 - 2 \cdot cos^{2} C + 1 2 \cdot cos^{2} A + 2 \cdot cos^{2} B - 2 \cdot cos^{2} C - 1 - 1 + 1 2 \cdot cos^{2} A + 2 \cdot cos^{2} B - 2 \cdot cos^{2} C - 1 2 \cdot cos^{2} A + 2 \cdot cos^{2} B - 2 \cdot (1 - sin^{2} A - sin^{2} B + 2 \cdot sin A \cdot sin B \cdot cos C) - 1 2 \cdot cos^{2} A + 2 \cdot cos^{2} B - 2 + 2 \cdot sin^{2} A + 2 \cdot sin^{2} B - 4 \cdot sin A \cdot sin B \cdot cos C - 1 - 2 - 1 + 2 \cdot cos^{2} A + 2 \cdot sin^{2} A + 2 \cdot cos^{2} B + 2 \cdot sin^{2} B - 4 \cdot sin A \cdot sin B \cdot cos C - 3 + 2 \cdot (cos^{2} A + sin^{2} A) + 2 \cdot (cos^{2} B + sin^{2} B) - 4 \cdot sin A \cdot sin B \cdot cos C - 3 + 2 \cdot (1) + 2 \cdot (1) - 4 \cdot sin A \cdot sin B \cdot cos C - 3 + 2 + 2 - 4 \cdot sin A \cdot sin B \cdot cos C - 3 + 4 - 4 \cdot sin A \cdot sin B \cdot cos C 1 - 4 \cdot sin A \cdot sin B \cdot cos C$