Pertanyaan

7. Jika m tan ( θ − 3 0 ∘ ) = n tan ( θ + 1 2 ∘ ) , tunjukkan bahwa cos 2 θ = 2 ( m − n ) m + n .

7. Jika $m tan (θ - 3 0^{\circ}) = n tan (θ + 1 2^{\circ})$ , tunjukkan bahwa $cos 2 θ = \frac{m + n}{2 ( m - n )}$ . $\;$

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

S. Nur

Master Teacher

Jawaban terverifikasi

Jawaban

telah ditunjukkan bahwa cos 2 θ = 2 ( m − n ) m + n .

telah ditunjukkan bahwa

cos 2 θ = \frac{m + n}{2 ( m - n )}

. $\;$

Pembahasan

Perhatian! Persamaan pada soal diralat menjadi: m tan ( θ − 3 0 ∘ ) = n tan ( θ + 12 0 ∘ ) Ingat! tan ( 9 0 ∘ + α ) = − cot α cot α = tan α 1 cos ( α + β ) = cos α ⋅ cos β − sin α ⋅ sin β cos ( α − β ) = cos α ⋅ cos β + sin α ⋅ sin β 1 = a a , dengan a  = 0 cos ( − 6 0 ∘ ) = 2 1 Oleh karena itu, dapat diperoleh: m tan ( θ − 3 0 ∘ ) m tan ( θ − 3 0 ∘ ) m tan ( θ − 3 0 ∘ ) m tan ( θ − 3 0 ∘ ) m tan ( θ − 3 0 ∘ ) tan ( θ − 3 0 ∘ ) ⋅ tan ( θ + 3 0 ∘ ) c o s ( θ − 3 0 ∘ ) s i n ( θ − 3 0 ∘ ) ⋅ c o s ( θ + 3 0 ∘ ) s i n ( θ + 3 0 ∘ ) c o s ( θ − 3 0 ∘ ) ⋅ c o s ( θ + 3 0 ∘ ) s i n ( θ − 3 0 ∘ ) ⋅ s i n ( θ + 3 0 ∘ ) = = = = = = = = n tan ( θ + 12 0 ∘ ) n tan ( 9 0 ∘ + ( θ + 3 0 ∘ ) ) n ( − cot ( θ + 3 0 ∘ ) ) n ( t a n ( θ + 3 0 ∘ ) − 1 ) t a n ( θ + 3 0 ∘ ) − n m − n m − n m − n Misalkan A = θ − 3 0 ∘ dan B = θ + 3 0 ∘ , maka dapat diperoleh: c o s ( A ) ⋅ c o s ( B ) s i n ( A ) ⋅ s i n ( B ) = m − n ( 1 ) dan saat kedua ruas pada (1) dikurangdengan 1, diperoleh: c o s ( A ) ⋅ c o s ( B ) s i n ( A ) ⋅ s i n ( B ) − 1 c o s ( A ) ⋅ c o s ( B ) s i n ( A ) ⋅ s i n ( B ) − c o s ( A ) ⋅ c o s ( B ) c o s ( A ) ⋅ c o s ( B ) c o s ( A ) ⋅ c o s ( B ) s i n ( A ) ⋅ s i n ( B ) − c o s ( A ) ⋅ c o s ( B ) c o s ( A ) ⋅ c o s ( B ) − c o s ( A ) ⋅ c o s ( B ) + s i n ( A ) ⋅ s i n ( B ) c o s ( A ) ⋅ c o s ( B ) − ( c o s ( A ) ⋅ c o s ( B ) − s i n ( A ) ⋅ s i n ( B ) ) c o s ( A ) ⋅ c o s ( B ) c o s ( A + B ) c o s ( θ − 3 0 ∘ ) ⋅ c o s ( θ + 3 0 ∘ ) c o s ( θ − 3 0 ∘ + θ + 3 0 ∘ ) c o s ( θ − 3 0 ∘ ) ⋅ c o s ( θ + 3 0 ∘ ) c o s ( 2 θ ) = = = = = = = = m − n − 1 m − n − m m m − n − m m − m − n m − ( m + n ) m m + n karena A = θ − 3 0 ∘ B = θ + 3 0 ∘ maka m m + n m m + n ( 2 ) dan saat kedua ruas pada (1) ditambahdengan 1, diperoleh: c o s ( A ) ⋅ c o s ( B ) s i n ( A ) ⋅ s i n ( B ) + 1 c o s ( A ) ⋅ c o s ( B ) s i n ( A ) ⋅ s i n ( B ) + c o s ( A ) ⋅ c o s ( B ) c o s ( A ) ⋅ c o s ( B ) c o s ( A ) ⋅ c o s ( B ) s i n ( A ) ⋅ s i n ( B ) + c o s ( A ) ⋅ c o s ( B ) c o s ( A ) ⋅ c o s ( B ) c o s ( A ) ⋅ c o s ( B ) + s i n ( A ) ⋅ s i n ( B ) c o s ( A ) ⋅ c o s ( B ) c o s ( A − B ) c o s ( θ − 3 0 ∘ ) ⋅ c o s ( θ + 3 0 ∘ ) c o s ( θ − 3 0 ∘ − ( θ + 3 0 ∘ ) ) c o s ( θ − 3 0 ∘ ) ⋅ c o s ( θ + 3 0 ∘ ) c o s ( θ − 3 0 ∘ − θ − 3 0 ∘ ) c o s ( θ − 3 0 ∘ ) ⋅ c o s ( θ + 3 0 ∘ ) c o s ( − 6 0 ∘ ) = = = = = = = = m − n + 1 m − n + m m m − n + m m m − n m m − n karena A = θ − 3 0 ∘ B = θ + 3 0 ∘ maka m m − n m m − n m m − n ( 3 ) Dengan membagi (2) dengan (3), diperoleh c o s ( θ − 3 0 ∘ ) ⋅ c o s ( θ + 3 0 ∘ ) c o s ( − 6 0 ∘ ) c o s ( θ − 3 0 ∘ ) ⋅ c o s ( θ + 3 0 ∘ ) c o s 2 θ c o s ( − 6 0 ∘ ) c o s 2 θ cos 2 θ = = = m m − n m m + n m − n m + n cos ( − 6 0 ∘ ) ⋅ m − n m + n dan karena cos ( − 6 0 ∘ ) = 2 1 , maka cos 2 θ cos 2 θ = = 2 1 ⋅ m − n m + n 2 ( m − n ) m + n Dengan demikian, telah ditunjukkan bahwa cos 2 θ = 2 ( m − n ) m + n .

Perhatian!
Persamaan pada soal diralat menjadi: $m tan (θ - 3 0^{\circ}) = n tan (θ + 12 0^{\circ})$

Ingat!

$tan (9 0^{\circ} + α) = - cot α$
$cot α = \frac{1}{tan α}$
$cos (α + β) = cos α \cdot cos β - sin α \cdot sin β$
$cos (α - β) = cos α \cdot cos β + sin α \cdot sin β$
$1 = \frac{a}{a}$ , dengan $a \neq = 0$
$cos (- 6 0^{\circ}) = \frac{1}{2}$

Oleh karena itu, dapat diperoleh:

$m tan (θ - 3 0^{\circ}) m tan (θ - 3 0^{\circ}) m tan (θ - 3 0^{\circ}) m tan (θ - 3 0^{\circ}) m tan (θ - 3 0^{\circ}) tan (θ - 3 0^{\circ}) \cdot tan (θ + 3 0^{\circ}) \frac{s i n ( θ - 3 0 ^{\circ} )}{c o s ( θ - 3 0 ^{\circ} )} \cdot \frac{s i n ( θ + 3 0 ^{\circ} )}{c o s ( θ + 3 0 ^{\circ} )} \frac{s i n ( θ - 3 0 ^{\circ} ) \cdot s i n ( θ + 3 0 ^{\circ} )}{c o s ( θ - 3 0 ^{\circ} ) \cdot c o s ( θ + 3 0 ^{\circ} )} = = = = = = = = n tan (θ + 12 0^{\circ}) n tan (9 0^{\circ} + (θ + 3 0^{\circ})) n (- cot (θ + 3 0^{\circ})) n (\frac{- 1}{t a n ( θ + 3 0 ^{\circ} )}) \frac{- n}{t a n ( θ + 3 0 ^{\circ} )} \frac{- n}{m} \frac{- n}{m} \frac{- n}{m}$

Misalkan $A = θ - 3 0^{\circ}$ dan $B = θ + 3 0^{\circ}$ , maka dapat diperoleh:

$\frac{s i n ( A ) \cdot s i n ( B )}{c o s ( A ) \cdot c o s ( B )} = \frac{- n}{m} (1)$

dan saat kedua ruas pada (1) dikurang dengan 1, diperoleh:

$\frac{s i n ( A ) \cdot s i n ( B )}{c o s ( A ) \cdot c o s ( B )} - 1 \frac{s i n ( A ) \cdot s i n ( B )}{c o s ( A ) \cdot c o s ( B )} - \frac{c o s ( A ) \cdot c o s ( B )}{c o s ( A ) \cdot c o s ( B )} \frac{s i n ( A ) \cdot s i n ( B ) - c o s ( A ) \cdot c o s ( B )}{c o s ( A ) \cdot c o s ( B )} \frac{- c o s ( A ) \cdot c o s ( B ) + s i n ( A ) \cdot s i n ( B )}{c o s ( A ) \cdot c o s ( B )} \frac{- ( c o s ( A ) \cdot c o s ( B ) - s i n ( A ) \cdot s i n ( B ) )}{c o s ( A ) \cdot c o s ( B )} \frac{c o s ( A + B )}{c o s ( A ) \cdot c o s ( B )} \frac{c o s ( θ - 3 0 ^{\circ} + θ + 3 0 ^{\circ} )}{c o s ( θ - 3 0 ^{\circ} ) \cdot c o s ( θ + 3 0 ^{\circ} )} \frac{c o s ( 2 θ )}{c o s ( θ - 3 0 ^{\circ} ) \cdot c o s ( θ + 3 0 ^{\circ} )} = = = = = = = = \frac{- n}{m} - 1 \frac{- n}{m} - \frac{m}{m} \frac{- n - m}{m} \frac{- m - n}{m} \frac{- ( m + n )}{m} \frac{m + n}{m}^{karena A = θ - 3 0^{\circ} B = θ + 3 0^{\circ}}^{maka} \frac{m + n}{m} \frac{m + n}{m} (2)$

dan saat kedua ruas pada (1) ditambah dengan 1, diperoleh:

$\frac{s i n ( A ) \cdot s i n ( B )}{c o s ( A ) \cdot c o s ( B )} + 1 \frac{s i n ( A ) \cdot s i n ( B )}{c o s ( A ) \cdot c o s ( B )} + \frac{c o s ( A ) \cdot c o s ( B )}{c o s ( A ) \cdot c o s ( B )} \frac{s i n ( A ) \cdot s i n ( B ) + c o s ( A ) \cdot c o s ( B )}{c o s ( A ) \cdot c o s ( B )} \frac{c o s ( A ) \cdot c o s ( B ) + s i n ( A ) \cdot s i n ( B )}{c o s ( A ) \cdot c o s ( B )} \frac{c o s ( A - B )}{c o s ( A ) \cdot c o s ( B )} \frac{c o s ( θ - 3 0 ^{\circ} - ( θ + 3 0 ^{\circ} ) )}{c o s ( θ - 3 0 ^{\circ} ) \cdot c o s ( θ + 3 0 ^{\circ} )} \frac{c o s ( θ - 3 0 ^{\circ} - θ - 3 0 ^{\circ} )}{c o s ( θ - 3 0 ^{\circ} ) \cdot c o s ( θ + 3 0 ^{\circ} )} \frac{c o s ( - 6 0 ^{\circ} )}{c o s ( θ - 3 0 ^{\circ} ) \cdot c o s ( θ + 3 0 ^{\circ} )} = = = = = = = = \frac{- n}{m} + 1 \frac{- n}{m} + \frac{m}{m} \frac{- n + m}{m} \frac{m - n}{m} \frac{m - n}{m}^{karena A = θ - 3 0^{\circ} B = θ + 3 0^{\circ}}^{maka} \frac{m - n}{m} \frac{m - n}{m} \frac{m - n}{m} (3)$

Dengan membagi (2) dengan (3), diperoleh

$\frac{\frac{c o s 2 θ}{c o s ( θ - 3 0 ^{\circ} ) \cdot c o s ( θ + 3 0 ^{\circ} )}}{\frac{c o s ( - 6 0 ^{\circ} )}{c o s ( θ - 3 0 ^{\circ} ) \cdot c o s ( θ + 3 0 ^{\circ} )}} \frac{c o s 2 θ}{c o s ( - 6 0 ^{\circ} )} cos 2 θ = = = \frac{\frac{m + n}{m}}{\frac{m - n}{m}} \frac{m + n}{m - n} cos (- 6 0^{\circ}) \cdot \frac{m + n}{m - n}$