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Tanpa menggunakan tabel matematika matematika maupun kalkulator, tentukan nilai dari: c. tan 14 2 ∘ 3 0 ′

Tanpa menggunakan tabel matematika matematika maupun kalkulator, tentukan nilai dari:

c. $tan 14 2^{\circ} 3 0^{'}$

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Pembahasan

Ingat kembali rumus: tan ( 90 − a ) = − co tan a cos A = cos 2 2 A − sin 2 2 A tan ( A + B ) = 1 − tan A ⋅ tan B tan A + tan B sin ( A − B ) = sin A cos B − cos A sin B cos ( A − B ) = cos A cos B + sin A sin B sin 2 A + cos 2 A = 1 Sehingga diperoleh perhitungan: tan 14 2 ∘ 3 0 ′ = = = = = = = = = = = = = = = = = = = = = = = = = = = tan 142 , 5 ∘ tan ( 90 + 52 , 5 ) ∘ − cotan 52 , 5 ∘ t a n 52 , 5 ∘ − 1 t a n ( 4 5 ∘ + 7 , 5 ∘ ) − 1 1 − tan 4 5 ∘ ⋅ tan 7 , 5 ∘ tan 4 5 ∘ + tan 7 , 5 ∘ − 1 1 − 1 ⋅ tan 7 , 5 ∘ 1 + tan 7 , 5 ∘ − 1 − 1 + t a n 7 , 5 ∘ 1 − t a n 7 , 5 ∘ − 1 + cos 7 , 5 ∘ sin 7 , 5 ∘ 1 − cos 7 , 5 ∘ sin 7 , 5 ∘ − cos 7 , 5 ∘ cos 7 , 5 ∘ + sin 7 , 5 ∘ cos 7 , 5 ∘ cos 7 , 5 ∘ − sin 7 , 5 ∘ − c o s 7 , 5 ∘ + s i n 7 , 5 ∘ c o s 7 , 5 ∘ − s i n 7 , 5 ∘ , misalkan A = 7 , 5 ∘ − c o s A + s i n A c o s A − s i n A × c o s A − s i n A c o s A − s i n A − c o s 2 A − s i n 2 A ( c o s A − s i n A ) 2 − c o s 2 A − s i n 2 A 1 − 2 s i n A c o s A − c o s 2 A 1 − s i n 2 A − c o s 2 ⋅ 7 , 5 ∘ 1 − s i n 2 ⋅ 7 , 5 ∘ − c o s 1 5 ∘ 1 − s i n 1 5 ∘ − c o s ( 4 5 ∘ − 3 0 ∘ ) 1 − s i n ( 4 5 ∘ − 3 0 ∘ ) − c o s 4 5 ∘ c o s 3 0 ∘ + s i n 4 5 ∘ s i n 3 0 ∘ 1 − ( s i n 4 5 ∘ c o s 3 0 ∘ − c o s 4 5 ∘ s i n 3 0 ∘ ) − 2 2 3 + 1 1 − 2 2 3 − 1 − 3 + 1 2 2 − 3 + 1 − 3 + 1 2 2 − 3 + 1 × 3 − 1 3 − 1 − 3 − 1 ( 2 2 − 3 + 1 ) ( 3 − 1 ) − 2 ( 2 2 ( 3 − 1 ) − ( 3 − 1 ) 2 ) − [ 2 ( 3 − 1 ) − ( 2 − 3 ) ] − 6 + 2 + 2 − 3 2 + 2 − 3 − 6 Jadi, nilai dari tan 14 2 ∘ 3 0 ′ adalah 2 + 2 − 3 − 6 .

Ingat kembali rumus:

$tan (90 - a) = - co tan a$

$cos A = cos^{2} \frac{A}{2} - sin^{2} \frac{A}{2}$

$tan (A + B) = \frac{tan A + tan B}{1 - tan A \cdot tan B}$

$sin (A - B) = sin A cos B - cos A sin B$

$cos (A - B) = cos A cos B + sin A sin B$

$sin^{2} A + cos^{2} A = 1$

Sehingga diperoleh perhitungan:

$tan 14 2^{\circ} 3 0^{'} = = = = = = = = = = = = = = = = = = = = = = = = = = = tan 142, 5^{\circ} tan (90 + 52, 5)^{\circ} - cotan 52, 5^{\circ} \frac{- 1}{t a n 52 , 5 ^{\circ}} \frac{- 1}{t a n ( 4 5 ^{\circ} + 7 , 5 ^{\circ} )} \frac{- 1}{\frac{tan 4 5 ^{\circ} + tan 7 , 5 ^{\circ}}{1 - tan 4 5 ^{\circ} \cdot tan 7 , 5 ^{\circ}}} \frac{- 1}{\frac{1 + tan 7 , 5 ^{\circ}}{1 - 1 \cdot tan 7 , 5 ^{\circ}}} - \frac{1 - t a n 7 , 5 ^{\circ}}{1 + t a n 7 , 5 ^{\circ}} - \frac{1 - \frac{sin 7 , 5 ^{\circ}}{cos 7 , 5 ^{\circ}}}{1 + \frac{sin 7 , 5 ^{\circ}}{cos 7 , 5 ^{\circ}}} - \frac{\frac{cos 7 , 5 ^{\circ} - sin 7 , 5 ^{\circ}}{cos 7 , 5 ^{\circ}}}{\frac{cos 7 , 5 ^{\circ} + sin 7 , 5 ^{\circ}}{cos 7 , 5 ^{\circ}}} - \frac{c o s 7 , 5 ^{\circ} - s i n 7 , 5 ^{\circ}}{c o s 7 , 5 ^{\circ} + s i n 7 , 5 ^{\circ}}, misalkan A = 7, 5^{\circ} - \frac{c o s A - s i n A}{c o s A + s i n A} \times \frac{c o s A - s i n A}{c o s A - s i n A} - \frac{( c o s A - s i n A ) ^{2}}{c o s ^{2} A - s i n ^{2} A} - \frac{1 - 2 s i n A c o s A}{c o s ^{2} A - s i n ^{2} A} - \frac{1 - s i n 2 A}{c o s 2 A} - \frac{1 - s i n 2 \cdot 7 , 5 ^{\circ}}{c o s 2 \cdot 7 , 5 ^{\circ}} - \frac{1 - s i n 1 5 ^{\circ}}{c o s 1 5 ^{\circ}} - \frac{1 - s i n ( 4 5 ^{\circ} - 3 0 ^{\circ} )}{c o s ( 4 5 ^{\circ} - 3 0 ^{\circ} )} - \frac{1 - ( s i n 4 5 ^{\circ} c o s 3 0 ^{\circ} - c o s 4 5 ^{\circ} s i n 3 0 ^{\circ} )}{c o s 4 5 ^{\circ} c o s 3 0 ^{\circ} + s i n 4 5 ^{\circ} s i n 3 0 ^{\circ}} - \frac{1 - \frac{3 - 1}{2 2}}{\frac{3 + 1}{2 2}} - \frac{2 2 - 3 + 1}{3 + 1} - \frac{2 2 - 3 + 1}{3 + 1} \times \frac{3 - 1}{3 - 1} - \frac{( 2 2 - 3 + 1 ) ( 3 - 1 )}{3 - 1} - \frac{( 2 2 ( 3 - 1 ) - ( 3 - 1 ) ^{2} )}{2} - [2 (3 - 1) - (2 - 3)] - 6 + 2 + 2 - 3 2 + 2 - 3 - 6$