Pertanyaan

1. Jika tan β = 1 + tan α ⋅ tan γ tan α + tan γ , tunjukkan bahwa: sin 2 β = 1 + sin 2 α ⋅ sin 2 γ sin 2 α + sin 2 γ .

1. Jika $tan β = \frac{tan α + tan γ}{1 + tan α \cdot tan γ}$ , tunjukkan bahwa: $sin 2 β = \frac{sin 2 α + sin 2 γ}{1 + sin 2 α \cdot sin 2 γ}$ . $\;$

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Habis dalam

Klaim

I. Ridha

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

telah ditunjukan bahwa sin 2 β = 1 + sin 2 α ⋅ sin 2 γ sin 2 α + sin 2 γ .

telah ditunjukan bahwa

sin 2 β = \frac{sin 2 α + sin 2 γ}{1 + sin 2 α \cdot sin 2 γ}

. $\;$

Pembahasan

Ingat kembali: tan A = cos A sin A sin ( A + B ) = sin A ⋅ cos B + cos A ⋅ sin B sin ( A − B ) = sin A ⋅ cos B − cos A ⋅ sin B cos ( A + B ) = cos A ⋅ cos B − sin A ⋅ sin B cos ( A − B ) = cos A ⋅ cos B + sin A ⋅ sin B sin 2 A = 1 + tan 2 A 2 ⋅ tan A 2 ⋅ sin A ⋅ cos B = sin ( A + B ) + sin ( A − B ) 2 ⋅ cos 2 A = 1 + cos 2 A 2 ⋅ sin 2 A = 1 − cos 2 A Dengan menggunakan rumus-rumus di atas, dapat diperoleh tan β = = = = = = 1 + t a n α ⋅ t a n γ t a n α + t a n γ 1 + c o s α s i n α ⋅ c o s γ s i n γ c o s α s i n α + c o s γ s i n γ c o s α ⋅ c o s γ c o s α ⋅ c o s γ + c o s α ⋅ c o s γ s i n α ⋅ s i n γ c o s α ⋅ c o s γ s i n α ⋅ c o s γ + c o s α ⋅ c o s γ c o s α ⋅ s i n γ c o s α ⋅ c o s γ c o s α ⋅ c o s γ + s i n α ⋅ s i n γ c o s α ⋅ c o s γ s i n α ⋅ c o s γ + c o s α ⋅ s i n γ c o s α ⋅ c o s γ + s i n α ⋅ s i n γ s i n α ⋅ c o s γ + c o s α ⋅ s i n γ c o s ( α − γ ) s i n ( α + γ ) sehingga diperoleh sin 2 β sin 2 β = = = = = = = = = = = = = = 1 + t a n 2 β 2 ⋅ t a n β 1 + ( c o s ( α − γ ) s i n ( α + γ ) ) 2 2 ⋅ c o s ( α − γ ) s i n ( α + γ ) c o s 2 ( α − γ ) c o s 2 ( α − γ ) + c o s 2 ( α − γ ) s i n 2 ( α + γ ) 2 ⋅ c o s ( α − γ ) s i n ( α + γ ) c o s 2 ( α − γ ) c o s 2 ( α − γ ) + s i n 2 ( α + γ ) 2 ⋅ c o s ( α − γ ) s i n ( α + γ ) c o s ( α − γ ) 1 2 ⋅ s i n ( α + γ ) × c o s 2 ( α − γ ) + s i n 2 ( α + γ ) c o s 2 ( α − γ ) c o s ( α − γ ) c o s 2 ( α − γ ) + s i n 2 ( α + γ ) 2 ⋅ s i n ( α + γ ) ⋅ c o s ( α − γ ) c o s 2 ( α − γ ) + s i n 2 ( α + γ ) s i n ( α + γ + α − γ ) + s i n ( α + γ − ( α − γ ) ) c o s 2 ( α − γ ) + s i n 2 ( α + γ ) s i n 2 α + s i n 2 γ 2 ⋅ c o s 2 ( α − γ ) + 2 ⋅ s i n 2 ( α + γ ) 2 ⋅ ( s i n 2 α + s i n 2 γ ) 1 + c o s 2 ( α − γ ) + 1 − c o s 2 ( α + γ ) 2 ⋅ ( s i n 2 α + s i n 2 γ ) 2 + c o s ( 2 α − 2 γ ) − c o s ( 2 α + 2 γ ) 2 ⋅ ( s i n 2 α + s i n 2 γ ) Olehkarena c o s ( 2 α − 2 γ ) − c o s ( 2 α + 2 γ ) = c o s 2 α ⋅ c o s 2 γ + s i n 2 α ⋅ s i n 2 γ − ( c o s 2 α ⋅ c o s 2 γ − s i n 2 α ⋅ s i n 2 γ ) = c o s 2 α ⋅ c o s 2 γ + s i n 2 α ⋅ s i n 2 γ − c o s 2 α ⋅ c o s 2 γ + s i n 2 α ⋅ s i n 2 γ ) = c o s 2 α ⋅ c o s 2 γ − c o s 2 α ⋅ c o s 2 γ + s i n 2 α ⋅ s i n 2 γ + s i n 2 α ⋅ s i n 2 γ = 2 ⋅ s i n 2 α ⋅ s i n 2 γ makadiperoleh 2 + 2 ⋅ s i n 2 α ⋅ s i n 2 γ 2 ⋅ ( s i n 2 α + s i n 2 γ ) 2 ⋅ ( 1 + s i n 2 α ⋅ s i n 2 γ ) 2 ⋅ ( s i n 2 α + s i n 2 γ ) 1 + s i n 2 α ⋅ s i n 2 γ s i n 2 α + s i n 2 γ Dengan demikian, telah ditunjukan bahwa sin 2 β = 1 + sin 2 α ⋅ sin 2 γ sin 2 α + sin 2 γ .

Ingat kembali:

$tan A = \frac{sin A}{cos A}$
$sin (A + B) = sin A \cdot cos B + cos A \cdot sin B$
$sin (A - B) = sin A \cdot cos B - cos A \cdot sin B$
$cos (A + B) = cos A \cdot cos B - sin A \cdot sin B$
$cos (A - B) = cos A \cdot cos B + sin A \cdot sin B$
$sin 2 A = \frac{2 \cdot tan A}{1 + tan ^{2} A}$
$2 \cdot sin A \cdot cos B = sin (A + B) + sin (A - B)$
$2 \cdot cos^{2} A = 1 + cos 2 A$
$2 \cdot sin^{2} A = 1 - cos 2 A$

Dengan menggunakan rumus-rumus di atas, dapat diperoleh

$tan β = = = = = = \frac{t a n α + t a n γ}{1 + t a n α \cdot t a n γ} \frac{\frac{s i n α}{c o s α} + \frac{s i n γ}{c o s γ}}{1 + \frac{s i n α}{c o s α} \cdot \frac{s i n γ}{c o s γ}} \frac{\frac{s i n α \cdot c o s γ}{c o s α \cdot c o s γ} + \frac{c o s α \cdot s i n γ}{c o s α \cdot c o s γ}}{\frac{c o s α \cdot c o s γ}{c o s α \cdot c o s γ} + \frac{s i n α \cdot s i n γ}{c o s α \cdot c o s γ}} \frac{\frac{s i n α \cdot c o s γ + c o s α \cdot s i n γ}{c o s α \cdot c o s γ}}{\frac{c o s α \cdot c o s γ + s i n α \cdot s i n γ}{c o s α \cdot c o s γ}} \frac{s i n α \cdot c o s γ + c o s α \cdot s i n γ}{c o s α \cdot c o s γ + s i n α \cdot s i n γ} \frac{s i n ( α + γ )}{c o s ( α - γ )}$

sehingga diperoleh

$sin 2 β sin 2 β = = = = = = = = = = = = = = \frac{2 \cdot t a n β}{1 + t a n ^{2} β} \frac{2 \cdot \frac{s i n ( α + γ )}{c o s ( α - γ )}}{1 + ( \frac{s i n ( α + γ )}{c o s ( α - γ )} ) ^{2}} \frac{2 \cdot \frac{s i n ( α + γ )}{c o s ( α - γ )}}{\frac{c o s ^{2} ( α - γ )}{c o s ^{2} ( α - γ )} + \frac{s i n ^{2} ( α + γ )}{c o s ^{2} ( α - γ )}} \frac{2 \cdot \frac{s i n ( α + γ )}{c o s ( α - γ )}}{\frac{c o s ^{2} ( α - γ ) + s i n ^{2} ( α + γ )}{c o s ^{2} ( α - γ )}} \frac{2 \cdot s i n ( α + γ )}{c o s ( α - γ ) ^{1}} \times \frac{c o s ^{2} ( α - γ ) ^{c o s (α - γ)}}{c o s ^{2} ( α - γ ) + s i n ^{2} ( α + γ )} \frac{2 \cdot s i n ( α + γ ) \cdot c o s ( α - γ )}{c o s ^{2} ( α - γ ) + s i n ^{2} ( α + γ )} \frac{s i n ( α + γ + α - γ ) + s i n ( α + γ - ( α - γ ) )}{c o s ^{2} ( α - γ ) + s i n ^{2} ( α + γ )} \frac{s i n 2 α + s i n 2 γ}{c o s ^{2} ( α - γ ) + s i n ^{2} ( α + γ )} \frac{2 \cdot ( s i n 2 α + s i n 2 γ )}{2 \cdot c o s ^{2} ( α - γ ) + 2 \cdot s i n ^{2} ( α + γ )} \frac{2 \cdot ( s i n 2 α + s i n 2 γ )}{1 + c o s 2 ( α - γ ) + 1 - c o s 2 ( α + γ )} \frac{2 \cdot ( s i n 2 α + s i n 2 γ )}{2 + c o s ( 2 α - 2 γ ) - c o s ( 2 α + 2 γ )}_{Oleh karena c o s (2 α - 2 γ) - c o s (2 α + 2 γ) = c o s 2 α \cdot c o s 2 γ + s i n 2 α \cdot s i n 2 γ - (c o s 2 α \cdot c o s 2 γ - s i n 2 α \cdot s i n 2 γ) = c o s 2 α \cdot c o s 2 γ + s i n 2 α \cdot s i n 2 γ - c o s 2 α \cdot c o s 2 γ + s i n 2 α \cdot s i n 2 γ) = c o s 2 α \cdot c o s 2 γ - c o s 2 α \cdot c o s 2 γ + s i n 2 α \cdot s i n 2 γ + s i n 2 α \cdot s i n 2 γ = 2 \cdot s i n 2 α \cdot s i n 2 γ maka diperoleh} \frac{2 \cdot ( s i n 2 α + s i n 2 γ )}{2 + 2 \cdot s i n 2 α \cdot s i n 2 γ} \frac{2 \cdot ( s i n 2 α + s i n 2 γ )}{2 \cdot ( 1 + s i n 2 α \cdot s i n 2 γ )} \frac{s i n 2 α + s i n 2 γ}{1 + s i n 2 α \cdot s i n 2 γ}$