Pertanyaan

Jika tan 2 θ − sin 2 θ = x dan tan 2 θ + sin 2 θ = y , tunjukkan bahwa: b. ( x 2 − y 2 ) 2 = 16 x y

Jika $tan 2 θ - sin 2 θ = x$ dan $tan 2 θ + sin 2 θ = y$ , tunjukkan bahwa:

b. $(x^{2} - y^{2})^{2} = 16 x y$

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Habis dalam

Klaim

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Pembahasan

Konsep: c o s A s i n A sin 2 A + cos 2 A = = tan A 1 Pembuktian: Pembuktian ruas kiri: = = = = = = = = = ( x 2 − y 2 ) 2 ( ( x + y ) ( x − y ) ) 2 ( ( tan 2 θ − sin 2 θ + tan 2 θ + sin 2 θ ) ( tan 2 θ − sin 2 θ − tan 2 θ − sin 2 θ )) 2 ( ( 2 tan 2 θ ) ( − 2 sin 2 θ ) ) 2 ( − 4 tan 2 θ sin 2 θ ) 2 ( − 4 c o s 2 θ s i n 2 θ sin 2 θ ) 2 ( − 4 c o s 2 θ s i n 2 2 θ ) 2 16 c o s 2 2 θ s i n 4 2 θ 16 c o s 2 2 θ s i n 2 2 θ ⋅ s i n 2 2 θ 16 tan 2 θ sin 2 θ Pembuktian ruas kanan: 16 x y = = = = = = = 16 ( tan 2 θ − sin 2 θ ) ( tan 2 θ + sin 2 θ ) 16 ( tan 2 2 θ − sin 2 2 θ ) 16 ( c o s 2 2 θ s i n 2 2 θ − sin 2 2 θ ) 16 ( c o s 2 2 θ s i n 2 2 θ − ( s i n 2 2 θ ) ( c o s 2 2 θ ) ) 16 ( c o s 2 2 θ s i n 2 2 θ ( 1 − c o s 2 2 θ ) ) 16 ( c o s 2 2 θ s i n 2 2 θ ( s i n 2 2 θ ) ) 16 tan 2 2 θ sin 2 2 θ Jadi, terbukti bahwa ruas kiri sama dengan ruas kanan ( x 2 − y 2 ) 2 = 16 x y .

Konsep:

$\frac{s i n A}{c o s A} sin^{2} A + cos^{2} A = = tan A 1$

Pembuktian:

Pembuktian ruas kiri:

$= = = = = = = = = (x^{2} - y^{2})^{2} ((x + y) (x - y))^{2} ((tan 2 θ - sin 2 θ + tan 2 θ + sin 2 θ) (tan 2 θ - sin 2 θ - tan 2 θ - sin 2 θ))^{2} ((2 tan 2 θ) (- 2 sin 2 θ))^{2} (- 4 tan 2 θ sin 2 θ)^{2} (- 4 \frac{s i n 2 θ}{c o s 2 θ} sin 2 θ)^{2} (- 4 \frac{s i n ^{2} 2 θ}{c o s 2 θ})^{2} 16 \frac{s i n ^{4} 2 θ}{c o s ^{2} 2 θ} 16 \frac{s i n ^{2} 2 θ \cdot s i n ^{2} 2 θ}{c o s ^{2} 2 θ} 16 tan^{2} θ sin^{2} θ$

Pembuktian ruas kanan:

$16 x y = = = = = = = 16 (tan 2 θ - sin 2 θ) (tan 2 θ + sin 2 θ) 16 (tan^{2} 2 θ - sin^{2} 2 θ) 16 (\frac{s i n ^{2} 2 θ}{c o s ^{2} 2 θ} - sin^{2} 2 θ) 16 (\frac{s i n ^{2} 2 θ - ( s i n ^{2} 2 θ ) ( c o s ^{2} 2 θ )}{c o s ^{2} 2 θ}) 16 (\frac{s i n ^{2} 2 θ ( 1 - c o s ^{2} 2 θ )}{c o s ^{2} 2 θ}) 16 (\frac{s i n ^{2} 2 θ ( s i n ^{2} 2 θ )}{c o s ^{2} 2 θ}) 16 tan^{2} 2 θ sin^{2} 2 θ$