Pertanyaan

Buktikanlah cos 0 ∘ ⋅ cos 1 ∘ 1 + cos 1 ∘ ⋅ cos 2 ∘ 1 + ... + cos 8 8 ∘ ⋅ cos 8 9 ∘ 1 = sin 2 1 ∘ cos 1 ∘

Buktikanlah

$\frac{1}{cos 0 ^{\circ} \cdot cos 1 ^{\circ}} + \frac{1}{cos 1 ^{\circ} \cdot cos 2 ^{\circ}} + ... + \frac{1}{cos 8 8 ^{\circ} \cdot cos 8 9 ^{\circ}} = \frac{cos 1 ^{\circ}}{sin ^{2} 1 ^{\circ}}$ $\;\;$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

terbukti bahwa cos 0 ∘ ⋅ cos 1 ∘ 1 + cos 1 ∘ ⋅ cos 2 ∘ 1 + ... + cos 8 8 ∘ ⋅ cos 8 9 ∘ 1 = sin 2 1 ∘ cos 1 ∘ .

terbukti bahwa

\frac{1}{cos 0 ^{\circ} \cdot cos 1 ^{\circ}} + \frac{1}{cos 1 ^{\circ} \cdot cos 2 ^{\circ}} + ... + \frac{1}{cos 8 8 ^{\circ} \cdot cos 8 9 ^{\circ}} = \frac{cos 1 ^{\circ}}{sin ^{2} 1 ^{\circ}}

. $\;\;$

Pembahasan

Ingat, Tangen sudut tan x = cos x sin x tan A − tan B tan A − tan B tan A − tan B = = = c o s A s i n A − c o s B s i n B c o s A c o s B s i n A c o s B − s i n B c o s A c o s A c o s B s i n ( A − B ) Berdasarkan rumus tersebut, diperoleh sebagai berikut Buktikan cos 0 ∘ ⋅ cos 1 ∘ 1 + cos 1 ∘ ⋅ cos 2 ∘ 1 + ... + cos 8 8 ∘ ⋅ cos 8 9 ∘ 1 = sin 2 1 ∘ cos 1 ∘ = = = = = = = = = = c o s 0 ∘ ⋅ c o s 1 ∘ 1 + c o s 1 ∘ ⋅ c o s 2 ∘ 1 + ... + c o s 8 8 ∘ ⋅ c o s 8 9 ∘ 1 s i n 1 ∘ 1 [ c o s 0 ∘ ⋅ c o s 1 ∘ s i n 1 ∘ + c o s 1 ∘ ⋅ c o s 2 ∘ s i n 1 ∘ + ... + c o s 8 8 ∘ ⋅ c o s 8 9 ∘ s i n 1 ∘ ] s i n 1 ∘ 1 [ c o s 0 ∘ ⋅ c o s 1 ∘ s i n 1 ∘ + c o s 1 ∘ ⋅ c o s 2 ∘ s i n 1 ∘ + ... + c o s 8 8 ∘ ⋅ c o s 8 9 ∘ s i n 1 ∘ ] s i n 1 ∘ 1 [ c o s 0 ∘ ⋅ c o s 1 ∘ s i n ( 1 − 0 ) ∘ + c o s 1 ∘ ⋅ c o s 2 ∘ s i n ( 2 − 1 ) ∘ + ... + c o s 8 8 ∘ ⋅ c o s 8 9 ∘ s i n ( 89 − 88 ) ∘ ] s i n 1 ∘ 1 [ ( tan 1 ∘ − tan 0 ∘ ) + ( tan 2 ∘ − tan 1 ∘ ) + ... + ( tan 8 9 ∘ − tan 8 8 ∘ ) ] s i n 1 ∘ 1 [ − tan 0 ∘ + tan 8 9 ∘ ] s i n 1 ∘ 1 [ tan 8 9 ∘ ] s i n 1 ∘ 1 [ tan ( 90 − 1 ) ∘ ] s i n 1 ∘ 1 [ co tan 1 ∘ ] s i n 2 1 ∘ c o s 1 ∘ Dengan demikian, terbukti bahwa cos 0 ∘ ⋅ cos 1 ∘ 1 + cos 1 ∘ ⋅ cos 2 ∘ 1 + ... + cos 8 8 ∘ ⋅ cos 8 9 ∘ 1 = sin 2 1 ∘ cos 1 ∘ .

Ingat,

Tangen sudut

$tan x = \frac{sin x}{cos x}$

$tan A - tan B tan A - tan B tan A - tan B = = = \frac{s i n A}{c o s A} - \frac{s i n B}{c o s B} \frac{s i n A c o s B - s i n B c o s A}{c o s A c o s B} \frac{s i n ( A - B )}{c o s A c o s B}$

Berdasarkan rumus tersebut, diperoleh sebagai berikut

Buktikan

$\frac{1}{cos 0 ^{\circ} \cdot cos 1 ^{\circ}} + \frac{1}{cos 1 ^{\circ} \cdot cos 2 ^{\circ}} + ... + \frac{1}{cos 8 8 ^{\circ} \cdot cos 8 9 ^{\circ}} = \frac{cos 1 ^{\circ}}{sin ^{2} 1 ^{\circ}}$

$= = = = = = = = = = \frac{1}{c o s 0 ^{\circ} \cdot c o s 1 ^{\circ}} + \frac{1}{c o s 1 ^{\circ} \cdot c o s 2 ^{\circ}} + ... + \frac{1}{c o s 8 8 ^{\circ} \cdot c o s 8 9 ^{\circ}} \frac{1}{s i n 1 ^{\circ}} [\frac{s i n 1 ^{\circ}}{c o s 0 ^{\circ} \cdot c o s 1 ^{\circ}} + \frac{s i n 1 ^{\circ}}{c o s 1 ^{\circ} \cdot c o s 2 ^{\circ}} + ... + \frac{s i n 1 ^{\circ}}{c o s 8 8 ^{\circ} \cdot c o s 8 9 ^{\circ}}] \frac{1}{s i n 1 ^{\circ}} [\frac{s i n 1 ^{\circ}}{c o s 0 ^{\circ} \cdot c o s 1 ^{\circ}} + \frac{s i n 1 ^{\circ}}{c o s 1 ^{\circ} \cdot c o s 2 ^{\circ}} + ... + \frac{s i n 1 ^{\circ}}{c o s 8 8 ^{\circ} \cdot c o s 8 9 ^{\circ}}] \frac{1}{s i n 1 ^{\circ}} [\frac{s i n ( 1 - 0 ) ^{\circ}}{c o s 0 ^{\circ} \cdot c o s 1 ^{\circ}} + \frac{s i n ( 2 - 1 ) ^{\circ}}{c o s 1 ^{\circ} \cdot c o s 2 ^{\circ}} + ... + \frac{s i n ( 89 - 88 ) ^{\circ}}{c o s 8 8 ^{\circ} \cdot c o s 8 9 ^{\circ}}] \frac{1}{s i n 1 ^{\circ}} [(tan 1^{\circ} - tan 0^{\circ}) + (tan 2^{\circ} - tan 1^{\circ}) + ... + (tan 8 9^{\circ} - tan 8 8^{\circ})] \frac{1}{s i n 1 ^{\circ}} [- tan 0^{\circ} + tan 8 9^{\circ}] \frac{1}{s i n 1 ^{\circ}} [tan 8 9^{\circ}] \frac{1}{s i n 1 ^{\circ}} [tan (90 - 1)^{\circ}] \frac{1}{s i n 1 ^{\circ}} [co tan 1^{\circ}] \frac{c o s 1 ^{\circ}}{s i n ^{2} 1 ^{\circ}}$

Dengan demikian, terbukti bahwa $\frac{1}{cos 0 ^{\circ} \cdot cos 1 ^{\circ}} + \frac{1}{cos 1 ^{\circ} \cdot cos 2 ^{\circ}} + ... + \frac{1}{cos 8 8 ^{\circ} \cdot cos 8 9 ^{\circ}} = \frac{cos 1 ^{\circ}}{sin ^{2} 1 ^{\circ}}$ . $\;\;$