Pertanyaan

Nilai 1 − tan 2 1 5 ∘ 2 3 tan 1 5 ∘ + 1 − tan 2 22 , 5 ∘ 2 tan 22 , 5 ∘ adalah ....

Nilai $\frac{2 3 tan 1 5 ^{\circ}}{1 - tan ^{2} 1 5 ^{\circ}} + \frac{2 tan 22 , 5 ^{\circ}}{1 - tan ^{2} 22 , 5 ^{\circ}}$ adalah ....

$4$
$2$
$\frac{1}{3} 3 + 1$
$1$
$\frac{1}{3} 3 - 1$

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Habis dalam

Klaim

L. Nikmah

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

Pembahasan

Ingat rumus tangen sudut paruh di bawah ini, tan 2 1 α = sin α 1 − cos α Berdasarkan rumus di atas, maka diperoleh perhitungan 1 − tan 2 1 5 ∘ 2 3 tan 1 5 ∘ + 1 − tan 2 22 , 5 ∘ 2 tan 22 , 5 ∘ = 1 − tan 2 ( 2 1 ⋅ 3 0 ∘ ) 2 3 tan ( 2 1 ⋅ 3 0 ∘ ) + 1 − tan 2 ( 2 1 ⋅ 4 5 ∘ ) 2 tan ( 2 1 ⋅ 4 5 ∘ ) = 1 − ( s i n 3 0 ∘ 1 − c o s 3 0 ∘ ) 2 2 3 ⋅ ( s i n 3 0 ∘ 1 − c o s 3 0 ∘ ) + 1 − ( s i n 4 5 ∘ 1 − c o s 4 5 ∘ ) 2 2 ( s i n 4 5 ∘ 1 − c o s 4 5 ∘ ) = 1 − ( 2 1 1 − 2 1 3 ) 2 2 3 ⋅ ( 2 1 1 − 2 1 3 ) + 1 − ( 2 1 2 1 − 2 1 2 ) 2 2 ( 2 1 2 1 − 2 1 2 ) = 1 − ( 2 1 1 − 2 1 3 ⋅ 2 2 ) 2 2 3 ⋅ ( 2 1 1 − 2 1 3 ⋅ 2 2 ) + 1 − ( 2 1 2 1 − 2 1 2 ⋅ 2 2 ) 2 2 ( 2 1 2 1 − 2 1 2 ⋅ 2 2 ) = 1 − ( 2 − 3 ) 2 2 3 ⋅ ( 2 − 3 ) + 1 − ( 2 − 1 ) 2 2 ( 2 − 1 ) = 1 − ( 4 − 4 3 + 3 ) 4 3 − 6 + 1 − ( 2 − 2 2 + 1 ) 2 2 − 2 = 1 − ( 7 − 4 3 ) 4 3 − 6 + 1 − ( 3 − 2 2 ) 2 2 − 2 = 1 − 7 + 4 3 4 3 − 6 + 1 − 3 + 2 2 2 2 − 2 = − 6 + 4 3 4 3 − 6 + − 2 + 2 2 2 2 − 2 = − 6 + 4 3 − 6 + 4 3 + − 2 + 2 2 − 2 + 2 2 = 1 + 1 = 2 Dengan demikian, nilai 1 − tan 2 1 5 ∘ 2 3 tan 1 5 ∘ + 1 − tan 2 22 , 5 ∘ 2 tan 22 , 5 ∘ adalah 2 . Jadi, jawaban yang benar adalah B.

Ingat rumus tangen sudut paruh di bawah ini,

$tan \frac{1}{2} α = \frac{1 - cos α}{sin α}$

Berdasarkan rumus di atas, maka diperoleh perhitungan

$\frac{2 3 tan 1 5 ^{\circ}}{1 - tan ^{2} 1 5 ^{\circ}} + \frac{2 tan 22 , 5 ^{\circ}}{1 - tan ^{2} 22 , 5 ^{\circ}} = \frac{2 3 tan ( \frac{1}{2} \cdot 3 0 ^{\circ} )}{1 - tan ^{2} ( \frac{1}{2} \cdot 3 0 ^{\circ} )} + \frac{2 tan ( \frac{1}{2} \cdot 4 5 ^{\circ} )}{1 - tan ^{2} ( \frac{1}{2} \cdot 4 5 ^{\circ} )} = \frac{2 3 \cdot ( \frac{1 - c o s 3 0 ^{\circ}}{s i n 3 0 ^{\circ}} )}{1 - ( \frac{1 - c o s 3 0 ^{\circ}}{s i n 3 0 ^{\circ}} ) ^{2}} + \frac{2 ( \frac{1 - c o s 4 5 ^{\circ}}{s i n 4 5 ^{\circ}} )}{1 - ( \frac{1 - c o s 4 5 ^{\circ}}{s i n 4 5 ^{\circ}} ) ^{2}} = \frac{2 3 \cdot ( \frac{1 - \frac{1}{2} 3}{\frac{1}{2}} )}{1 - ( \frac{1 - \frac{1}{2} 3}{\frac{1}{2}} ) ^{2}} + \frac{2 ( \frac{1 - \frac{1}{2} 2}{\frac{1}{2} 2} )}{1 - ( \frac{1 - \frac{1}{2} 2}{\frac{1}{2} 2} ) ^{2}} = \frac{2 3 \cdot ( \frac{1 - \frac{1}{2} 3}{\frac{1}{2}} \cdot \frac{2}{2} )}{1 - ( \frac{1 - \frac{1}{2} 3}{\frac{1}{2}} \cdot \frac{2}{2} ) ^{2}} + \frac{2 ( \frac{1 - \frac{1}{2} 2}{\frac{1}{2} 2} \cdot \frac{2}{2} )}{1 - ( \frac{1 - \frac{1}{2} 2}{\frac{1}{2} 2} \cdot \frac{2}{2} ) ^{2}} = \frac{2 3 \cdot ( 2 - 3 )}{1 - ( 2 - 3 ) ^{2}} + \frac{2 ( 2 - 1 )}{1 - ( 2 - 1 ) ^{2}} = \frac{4 3 - 6}{1 - ( 4 - 4 3 + 3 )} + \frac{2 2 - 2}{1 - ( 2 - 2 2 + 1 )} = \frac{4 3 - 6}{1 - ( 7 - 4 3 )} + \frac{2 2 - 2}{1 - ( 3 - 2 2 )} = \frac{4 3 - 6}{1 - 7 + 4 3} + \frac{2 2 - 2}{1 - 3 + 2 2} = \frac{4 3 - 6}{- 6 + 4 3} + \frac{2 2 - 2}{- 2 + 2 2} = \frac{- 6 + 4 3}{- 6 + 4 3} + \frac{- 2 + 2 2}{- 2 + 2 2} = 1 + 1 = 2$

Dengan demikian, nilai $\frac{2 3 tan 1 5 ^{\circ}}{1 - tan ^{2} 1 5 ^{\circ}} + \frac{2 tan 22 , 5 ^{\circ}}{1 - tan ^{2} 22 , 5 ^{\circ}}$ adalah $2$ .