Pertanyaan

Jika A + B + C = π , tunjukkan bahwa sin ( 2 A ) + sin ( 2 B ) + sin ( 2 C ) = 1 + 4 sin ( 4 π − A ) sin ( 4 π − B ) sin ( 4 π − C )

Jika $A + B + C = π$ , tunjukkan bahwa

$sin (\frac{A}{2}) + sin (\frac{B}{2}) + sin (\frac{C}{2}) = 1 + 4 sin (\frac{π - A}{4}) sin (\frac{π - B}{4}) sin (\frac{π - C}{4})$

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

S. Nur

Master Teacher

Jawaban terverifikasi

Pembahasan

Ingat bahwa : Rumus penjumlahan dan selisih trigonometri sin α + sin β = 2 sin 2 1 ( α + β ) cos 2 1 ( α − β ) cos α − cos β = − 2 sin 2 1 ( α + β ) sin 2 1 ( α − β ) Sudut berelasi I sin ( 9 0 ∘ − α ) sin ( − x ) = = cos α sin x Sudut rangkap pada cosinus cos 2 A = 1 − 2 sin 2 A Dari soal diketahui A + B + C C 2 C sin 2 C = = = = = π π − ( A + B ) 2 π − 2 ( A + B ) sin ( 2 π − 2 ( A + B ) ) cos ( 2 A + B ) Sehingga sin ( 2 A ) + sin ( 2 B ) + sin ( 2 C ) = ( sin ( 2 A ) + sin ( 2 B ) ) + sin ( 2 C ) = 2 sin 2 1 ( 2 A + 2 B ) cos 2 1 ( 2 A − 2 B ) + sin ( 2 π − 2 A + B ) = 2 sin ( 4 A + B ) cos ( 4 A − B ) + cos ( 2 A + B ) = 2 sin ( 4 A + B ) cos ( 4 A − B ) + 1 − 2 sin 2 ( 4 A + B ) = 1 + 2 sin ( 4 π − 4 C ) [ cos ( 4 A − B ) − sin ( 4 A + B ) ] = 1 + 2 sin ( 4 π − C ) [ cos ( 4 A − B ) − cos ( 2 π − ( 4 A + B ) ) ] = 1 + 2 sin ( 4 π − C ) [ cos ( 4 A − B ) − cos ( 4 2 π − 4 ( A + B ) ) ] = 1 + 2 sin ( 4 π − C ) [ − 2 sin 2 1 ( 4 A − B + ( 4 2 π − A − B ) ) sin 2 1 ( 4 A − B − ( 4 2 π − A − B ) ) ] = 1 + 2 sin ( 4 π − C ) [ − 2 sin 2 1 ( 4 2 π − 2 B ) sin 2 1 ( 4 2 A − 2 π ) ] = 1 + 2 sin ( 4 π − C ) [ − 2 sin ( 4 π − B ) ⋅ sin ( − 4 ( π − A ) ) ] = 1 + 2 sin ( 4 π − C ) [ − 2 sin ( 4 π − B ) ⋅ − sin ( 4 π − A ) ] = 1 + 2 sin ( 4 π − C ) [ 2 sin ( 4 π − B ) sin ( 4 π − A ) ] = 1 + 4 sin ( 4 π − A ) sin ( 4 π − B ) sin ( 4 π − C ) Dengan demikian benar bahwa S in A + sin B + sin C = 4 cos ( 2 A ) cos ( 2 B ) cos ( 2 C )

Ingat bahwa :

Rumus penjumlahan dan selisih trigonometri

$sin α + sin β = 2 sin \frac{1}{2} (α + β) cos \frac{1}{2} (α - β) cos α - cos β = - 2 sin \frac{1}{2} (α + β) sin \frac{1}{2} (α - β)$

Sudut berelasi I

$sin (9 0^{\circ} - α) sin (- x) = = cos α sin x$

Sudut rangkap pada cosinus

$cos 2 A = 1 - 2 sin^{2} A$

Dari soal diketahui

$A + B + C C \frac{C}{2} sin \frac{C}{2} = = = = = π π - (A + B) \frac{π}{2} - \frac{( A + B )}{2} sin (\frac{π}{2} - \frac{( A + B )}{2}) cos (\frac{A + B}{2})$

Sehingga

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