Pertanyaan

8. Buktikan bahwa: d. cos 2 θ = 1 + tan 2 θ 1 − tan 2 θ

8. Buktikan bahwa:

d. $cos 2 θ = \frac{1 - tan ^{2} θ}{1 + tan ^{2} θ}$

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Habis dalam

Klaim

I. Ridha

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

terbukti jika cos 2 θ = 1 + tan 2 θ 1 − tan 2 θ .

terbukti jika

cos 2 θ = \frac{1 - tan ^{2} θ}{1 + tan ^{2} θ}

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Pembahasan

Ingat kembali: rumus cosinus untuk sudut ganda: cos 2 α = cos 2 α − sin 2 α identitas trigonometri: tan 2 α + 1 = se c 2 α tan α = sec α 1 cos α = sec α 1 Oleh karena itu,dapat diperoleh: cos 2 θ = = = = = = = = = = = = cos 2 θ − sin 2 θ ( cos 2 θ − sin 2 θ ) ⋅ 1 ( cos 2 θ − sin 2 θ ) ⋅ ( c o s 2 θ 1 ⋅ 1 c o s 2 θ ) ( c o s 2 θ c o s 2 θ − s i n 2 θ ) ⋅ cos 2 θ ( c o s 2 θ c o s 2 θ − c o s 2 θ s i n 2 θ ) ⋅ cos 2 θ ( c o s 2 θ 1 c o s 2 θ 1 − c o s 2 θ s i n 2 θ ) ⋅ cos 2 θ ( 1 1 − ( c o s θ s i n θ ⋅ c o s θ s i n θ ) ) ⋅ cos θ ⋅ cos θ ( 1 − ( tan θ ⋅ tan θ ) ) ⋅ sec θ 1 ⋅ sec θ 1 ( 1 − tan 2 θ ) ⋅ sec 2 θ 1 ( 1 − tan 2 θ ) ⋅ 1 + t a n 2 θ 1 1 + t a n 2 θ ( 1 − t a n 2 θ ) ⋅ 1 1 + t a n 2 θ 1 − t a n 2 θ Dengan demikian, terbukti jika cos 2 θ = 1 + tan 2 θ 1 − tan 2 θ .

Ingat kembali:

rumus cosinus untuk sudut ganda: $cos 2 α = cos^{2} α - sin^{2} α$
identitas trigonometri: $tan^{2} α + 1 = se c^{2} α$
$tan α = \frac{1}{sec α}$
$cos α = \frac{1}{sec α}$

Oleh karena itu, dapat diperoleh:

$cos 2 θ = = = = = = = = = = = = cos^{2} θ - sin^{2} θ (cos^{2} θ - sin^{2} θ) \cdot 1 (cos^{2} θ - sin^{2} θ) \cdot (\frac{1}{c o s ^{2} θ} \cdot \frac{c o s ^{2} θ}{1}) (\frac{c o s ^{2} θ - s i n ^{2} θ}{c o s ^{2} θ}) \cdot cos^{2} θ (\frac{c o s ^{2} θ}{c o s ^{2} θ} - \frac{s i n ^{2} θ}{c o s ^{2} θ}) \cdot cos^{2} θ (\frac{c o s ^{2} θ ^{1}}{c o s ^{2} θ ^{1}} - \frac{s i n ^{2} θ}{c o s ^{2} θ}) \cdot cos^{2} θ (\frac{1}{1} - (\frac{s i n θ}{c o s θ} \cdot \frac{s i n θ}{c o s θ})) \cdot cos θ \cdot cos θ (1 - (tan θ \cdot tan θ)) \cdot \frac{1}{sec θ} \cdot \frac{1}{sec θ} (1 - tan^{2} θ) \cdot \frac{1}{sec ^{2} θ} (1 - tan^{2} θ) \cdot \frac{1}{1 + t a n ^{2} θ} \frac{( 1 - t a n ^{2} θ ) \cdot 1}{1 + t a n ^{2} θ} \frac{1 - t a n ^{2} θ}{1 + t a n ^{2} θ}$