Pertanyaan

Jika A + B + C = π , buktikan bahwa: a. sin A + sin B + sin C = 4 cos 2 A ⋅ cos 2 B ⋅ cos 2 C

Jika $A + B + C = π$ , buktikan bahwa:

a. $sin A + sin B + sin C = 4 cos \frac{A}{2} \cdot cos \frac{B}{2} \cdot cos \frac{C}{2}$

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Jawaban terverifikasi

Jawaban

terbukti bahwa, sin A + sin B + sin C = 4 cos 2 A ⋅ cos 2 B ⋅ cos 2 C .

terbukti bahwa,

sin A + sin B + sin C = 4 cos \frac{A}{2} \cdot cos \frac{B}{2} \cdot cos \frac{C}{2}

Pembahasan

Ingat kembali rumus: 1. sin ( 90 − x ) = cos x 2. sin ( 180 − x ) = sin x 3. sin 2 x = 2 sin x ⋅ cos x → sin x = 2 sin 2 x ⋅ cos 2 x 4. cos 2 x 1 + cos 2 x = = 2 cos 2 x − 1 2 cos 2 x − 1 5. sin ( A + B ) = sin A ⋅ cos B + cos A ⋅ sin B Dari soal diketahui: A + B + C A + B C = = = π π − C π − ( A + B ) Sehingga, sin C = = sin ( π − sin ( A + B ) ) sin ( A + B ) Maka diperoleh perhitungan: sin A + sin B + sin C = sin A + sin B + sin ( A + B ) = sin A + sin B + sin A ⋅ cos B + cos A ⋅ sin B = sin A + sin A ⋅ cos B + sin B + cos A ⋅ sin B = sin A ( 1 + cos B ) + sin B ( 1 + cos A ) = ( 2 sin 2 A ⋅ cos 2 A ) ( 2 cos 2 2 B ) + ( 2 sin 2 B ⋅ cos 2 B ) ( 2 cos 2 2 A ) = 4 ⋅ sin 2 A ⋅ cos 2 A ⋅ cos 2 2 B + 4 ⋅ sin 2 B ⋅ cos 2 B ⋅ cos 2 2 A = 4 cos 2 A ⋅ cos 2 B ( sin 2 A ⋅ cos 2 B + sin 2 B ⋅ cos 2 A ) = 4 cos 2 A ⋅ cos 2 B ⋅ sin ( 2 A + 2 B ) = 4 cos 2 A ⋅ cos 2 B ⋅ sin ( 2 A + 2 B ) = 4 cos 2 A ⋅ cos 2 B ⋅ sin 2 1 ( A + B ) = 4 cos 2 A ⋅ cos 2 B ⋅ sin 2 1 ( π − C ) = 4 cos 2 A ⋅ cos 2 B ⋅ sin ( 90 − 2 C ) = 4 cos 2 A ⋅ cos 2 B ⋅ cos 2 C terbukti Jadi, terbukti bahwa, sin A + sin B + sin C = 4 cos 2 A ⋅ cos 2 B ⋅ cos 2 C .

Ingat kembali rumus:

$1. sin (90 - x) = cos x 2. sin (180 - x) = sin x 3. sin 2 x = 2 sin x \cdot cos x \to sin x = 2 sin \frac{x}{2} \cdot cos \frac{x}{2} 4. cos 2 x 1 + cos 2 x = = 2 cos^{2} x - 1 2 cos^{2} x - 1 5. sin (A + B) = sin A \cdot cos B + cos A \cdot sin B$

Dari soal diketahui:

$A + B + C A + B C = = = π π - C π - (A + B)$

Sehingga,

$sin C = = sin (π - sin (A + B)) sin (A + B)$

Maka diperoleh perhitungan:

$sin A + sin B + sin C = sin A + sin B + sin (A + B) = sin A + sin B + sin A \cdot cos B + cos A \cdot sin B = sin A + sin A \cdot cos B + sin B + cos A \cdot sin B = sin A (1 + cos B) + sin B (1 + cos A) = (2 sin \frac{A}{2} \cdot cos \frac{A}{2}) (2 cos^{2} \frac{B}{2}) + (2 sin \frac{B}{2} \cdot cos \frac{B}{2}) (2 cos^{2} \frac{A}{2}) = 4 \cdot sin \frac{A}{2} \cdot cos \frac{A}{2} \cdot cos^{2} \frac{B}{2} + 4 \cdot sin \frac{B}{2} \cdot cos \frac{B}{2} \cdot cos^{2} \frac{A}{2} = 4 cos \frac{A}{2} \cdot cos \frac{B}{2} (sin \frac{A}{2} \cdot cos \frac{B}{2} + sin \frac{B}{2} \cdot cos \frac{A}{2}) = 4 cos \frac{A}{2} \cdot cos \frac{B}{2} \cdot sin (\frac{A}{2} + \frac{B}{2}) = 4 cos \frac{A}{2} \cdot cos \frac{B}{2} \cdot sin (\frac{A}{2} + \frac{B}{2}) = 4 cos \frac{A}{2} \cdot cos \frac{B}{2} \cdot sin \frac{1}{2} (A + B) = 4 cos \frac{A}{2} \cdot cos \frac{B}{2} \cdot sin \frac{1}{2} (π - C) = 4 cos \frac{A}{2} \cdot cos \frac{B}{2} \cdot sin (90 - \frac{C}{2}) = 4 cos \frac{A}{2} \cdot cos \frac{B}{2} \cdot cos \frac{C}{2} terbukti$