Pertanyaan

2. Jika sin ( θ + B ) sin ( θ + A ) = sin 2 B sin 2 A , buktikan bahwa: tan 2 θ = tan A ⋅ tan B .

2. Jika $\frac{sin ( θ + A )}{sin ( θ + B )} = \frac{sin 2 A}{sin 2 B}$ , buktikan bahwa: $tan^{2} θ = tan A \cdot tan B$ . $\;$

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Habis dalam

Klaim

I. Ridha

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

telah ditunjukan bahwa tan 2 θ = tan A ⋅ tan B .

telah ditunjukan bahwa

tan^{2} θ = tan A \cdot tan B

. $\;$

Pembahasan

Ingat kembali: sin ( α + β ) = sin α ⋅ cos β + cos α ⋅ sin β sin ( α − β ) = sin α ⋅ cos β − cos α ⋅ sin β sin 2 α = 2 ⋅ sin α ⋅ cos α tan α = cos α sin α Dengan menggunakan rumus-rumus di atas, dapat diperoleh s i n ( θ + B ) s i n ( θ + A ) ( s i n ( θ + B ) s i n ( θ + A ) ) 2 ( s i n ( θ + B ) ) 2 ( s i n ( θ + A ) ) 2 sin 2 B ⋅ ( sin ( θ + A ) ) 2 = = = = s i n 2 B s i n 2 A s i n 2 B s i n 2 A s i n 2 B s i n 2 A sin 2 A ⋅ ( sin ( θ + B ) ) 2 ( 1 ) Dengan menjabarkan ruas kiri pada (1), diperoleh: = = = = = = sin 2 B ⋅ ( sin ( θ + A ) ) 2 sin 2 B ⋅ ( sin θ ⋅ cos A + cos θ ⋅ sin A ) 2 sin 2 B ⋅ ( sin 2 θ ⋅ cos 2 A + cos 2 θ ⋅ sin 2 A + 2 ⋅ sin θ ⋅ cos A ⋅ cos θ ⋅ sin A ) sin 2 B ⋅ ( sin 2 θ ⋅ cos 2 A + cos 2 θ ⋅ sin 2 A + 2 ⋅ sin A ⋅ cos A ⋅ sin θ ⋅ cos θ ) sin 2 B ⋅ sin 2 θ ⋅ cos 2 A + sin 2 B ⋅ cos 2 θ ⋅ sin 2 A + sin 2 B ⋅ 2 ⋅ sin A ⋅ cos A ⋅ sin θ ⋅ cos θ 2 ⋅ sin B ⋅ cos B ⋅ sin 2 θ ⋅ cos 2 A + 2 ⋅ sin B ⋅ cos B ⋅ cos 2 θ ⋅ sin 2 A + 2 ⋅ sin B ⋅ cos B ⋅ 2 ⋅ sin A ⋅ cos A ⋅ sin θ ⋅ cos θ 2 ⋅ sin B ⋅ cos B ⋅ sin 2 θ ⋅ cos 2 A + 2 ⋅ sin B ⋅ cos B ⋅ cos 2 θ ⋅ sin 2 A + 4 ⋅ sin B ⋅ cos B ⋅ sin A ⋅ cos A ⋅ sin θ ⋅ cos θ ( 2 ) Dengan menjabarkan ruas kananpada (1), diperoleh: = = = = = = sin 2 A ⋅ ( sin ( θ + B ) ) 2 sin 2 A ( sin θ ⋅ cos B + cos θ ⋅ sin B ) 2 sin 2 A ⋅ ( sin 2 θ ⋅ cos 2 B + cos 2 θ ⋅ sin 2 B + 2 ⋅ sin θ ⋅ cos B ⋅ cos θ ⋅ sin B ) sin 2 A ⋅ ( sin 2 θ ⋅ cos 2 B + cos 2 θ ⋅ sin 2 B + 2 ⋅ sin B ⋅ cos B ⋅ sin θ ⋅ cos θ ) sin 2 A ⋅ sin 2 θ ⋅ cos 2 B + sin 2 B ⋅ cos 2 θ ⋅ sin 2 B + sin 2 A ⋅ 2 ⋅ sin B ⋅ cos B ⋅ sin θ ⋅ cos θ 2 ⋅ sin A ⋅ cos A ⋅ sin 2 θ ⋅ cos 2 B + 2 ⋅ sin A ⋅ cos A ⋅ cos 2 θ ⋅ sin 2 B + 2 ⋅ sin A ⋅ cos A ⋅ 2 ⋅ sin B ⋅ cos B ⋅ sin θ ⋅ cos θ 2 ⋅ sin A ⋅ cos A ⋅ sin 2 θ ⋅ cos 2 B + 2 ⋅ sin A ⋅ cos A ⋅ cos 2 θ ⋅ sin 2 B + 4 ⋅ sin A ⋅ cos A ⋅ sin B ⋅ cos B ⋅ sin θ ⋅ cos θ ( 3 ) Ubah (2) dan (3) pada (1) sehingga diperoleh: c o s 2 θ s i n 2 θ ( c o s θ s i n θ ) 2 ( c o s θ s i n θ ) 2 ( c o s θ s i n θ ) 2 ( c o s θ s i n θ ) 2 maka,diperoleh: ( tan θ ) 2 tan 2 θ = = = = = = = ( 2 ⋅ s i n B ⋅ c o s B ⋅ c o s 2 A − 2 ⋅ s i n A ⋅ c o s A ⋅ c o s 2 B ) ( 2 ⋅ s i n A ⋅ c o s A ⋅ s i n 2 B − 2 ⋅ s i n B ⋅ c o s B ⋅ s i n 2 A ) 2 ⋅ c o s A ⋅ c o s B ⋅ ( s i n B ⋅ c o s A − s i n A ⋅ c o s B ) 2 ⋅ s i n A ⋅ s i n B ⋅ ( c o s A ⋅ s i n B − c o s B ⋅ s i n A ) 2 ⋅ c o s A ⋅ c o s B ⋅ ( c o s A ⋅ s i n B − s i n A ⋅ c o s B ) 2 ⋅ s i n A ⋅ s i n B ⋅ ( c o s A ⋅ s i n B − s i n A ⋅ c o s B ) c o s A ⋅ c o s B s i n A ⋅ s i n B c o s A s i n A ⋅ c o s B s i n B karena t a n α = c o s α s i n α tan A ⋅ tan B tan A ⋅ tan B Dengan demikian, telah ditunjukan bahwa tan 2 θ = tan A ⋅ tan B .

Ingat kembali:

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

Dengan menggunakan rumus-rumus di atas, dapat diperoleh

$\frac{s i n ( θ + A )}{s i n ( θ + B )} (\frac{s i n ( θ + A )}{s i n ( θ + B )})^{2} \frac{( s i n ( θ + A ) ) ^{2}}{( s i n ( θ + B ) ) ^{2}} sin 2 B \cdot (sin (θ + A))^{2} = = = = \frac{s i n 2 A}{s i n 2 B} \frac{s i n 2 A}{s i n 2 B} \frac{s i n 2 A}{s i n 2 B} sin 2 A \cdot (sin (θ + B))^{2} (1)$

Dengan menjabarkan ruas kiri pada (1), diperoleh:

$= = = = = = sin 2 B \cdot (sin (θ + A))^{2} sin 2 B \cdot (sin θ \cdot cos A + cos θ \cdot sin A)^{2} sin 2 B \cdot (sin^{2} θ \cdot cos^{2} A + cos^{2} θ \cdot sin^{2} A + 2 \cdot sin θ \cdot cos A \cdot cos θ \cdot sin A) sin 2 B \cdot (sin^{2} θ \cdot cos^{2} A + cos^{2} θ \cdot sin^{2} A + 2 \cdot sin A \cdot cos A \cdot sin θ \cdot cos θ) sin 2 B \cdot sin^{2} θ \cdot cos^{2} A + sin 2 B \cdot cos^{2} θ \cdot sin^{2} A + sin 2 B \cdot 2 \cdot sin A \cdot cos A \cdot sin θ \cdot cos θ 2 \cdot sin B \cdot cos B \cdot sin^{2} θ \cdot cos^{2} A + 2 \cdot sin B \cdot cos B \cdot cos^{2} θ \cdot sin^{2} A + 2 \cdot sin B \cdot cos B \cdot 2 \cdot sin A \cdot cos A \cdot sin θ \cdot cos θ 2 \cdot sin B \cdot cos B \cdot sin^{2} θ \cdot cos^{2} A + 2 \cdot sin B \cdot cos B \cdot cos^{2} θ \cdot sin^{2} A + 4 \cdot sin B \cdot cos B \cdot sin A \cdot cos A \cdot sin θ \cdot cos θ (2)$

Dengan menjabarkan ruas kanan pada (1), diperoleh:

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

Ubah (2) dan (3) pada (1) sehingga diperoleh:

$\frac{s i n ^{2} θ}{c o s ^{2} θ} (\frac{s i n θ}{c o s θ})^{2} (\frac{s i n θ}{c o s θ})^{2} (\frac{s i n θ}{c o s θ})^{2} (\frac{s i n θ}{c o s θ})^{2}^{maka, diperoleh:} (tan θ)^{2} tan^{2} θ = = = = = = = \frac{( 2 \cdot s i n A \cdot c o s A \cdot s i n ^{2} B - 2 \cdot s i n B \cdot c o s B \cdot s i n ^{2} A )}{( 2 \cdot s i n B \cdot c o s B \cdot c o s ^{2} A - 2 \cdot s i n A \cdot c o s A \cdot c o s ^{2} B )} \frac{2 \cdot s i n A \cdot s i n B \cdot ( c o s A \cdot s i n B - c o s B \cdot s i n A )}{2 \cdot c o s A \cdot c o s B \cdot ( s i n B \cdot c o s A - s i n A \cdot c o s B )} \frac{2 \cdot s i n A \cdot s i n B \cdot ( c o s A \cdot s i n B - s i n A \cdot c o s B )}{2 \cdot c o s A \cdot c o s B \cdot ( c o s A \cdot s i n B - s i n A \cdot c o s B )} \frac{s i n A \cdot s i n B}{c o s A \cdot c o s B} \frac{s i n A}{c o s A} \cdot \frac{s i n B}{c o s B}^{karena t a n α = \frac{s i n α}{c o s α}} tan A \cdot tan B tan A \cdot tan B$