Jika sin (x-60)=cos (x+45) , maka tentukan nilai tg x !

tan x = (√3 + √2) / (1 + √2)

Pembahasan

Diberikan persamaan \sin(x - 60^{\circ}) = \cos(x + 45^{\circ}).
Kita tahu bahwa \cos \theta = \sin(90^{\circ} - \theta).
Maka, \cos(x + 45^{\circ}) = \sin(90^{\circ} - (x + 45^{\circ})) = \sin(90^{\circ} - x - 45^{\circ}) = \sin(45^{\circ} - x).

Jadi, persamaan menjadi \sin(x - 60^{\circ}) = \sin(45^{\circ} - x).
Ini berarti ada dua kemungkinan:

1. x - 60^{\circ} = 45^{\circ} - x + k \\cdot 360^{\circ}
   2x = 105^{\circ} + k \\cdot 360^{\circ}
   x = 52.5^{\circ} + k \\cdot 180^{\circ}

2. x - 60^{\circ} = 180^{\circ} - (45^{\circ} - x) + k \\cdot 360^{\circ}
   x - 60^{\circ} = 180^{\circ} - 45^{\circ} + x + k \\cdot 360^{\circ}
   x - 60^{\circ} = 135^{\circ} + x + k \\cdot 360^{\circ}
   -60^{\circ} = 135^{\circ} + k \\cdot 360^{\circ}
   Ini adalah kontradiksi, sehingga tidak ada solusi dari kasus kedua.

Sekarang kita perlu mencari nilai \tan x dari x = 52.5^{\circ} + k \\cdot 180^{\circ}.
Untuk k=0, x = 52.5^{\circ}.

Kita bisa menggunakan identitas \tan(A+B) atau \tan(A-B).
Misalnya, x = 52.5^{\circ} = (105/2)^{\circ} = (45+60)/2^{\circ} atau x = 105^{\circ} - 52.5^{\circ}.

Alternatif lain, gunakan identitas \sin A = \cos B \implies A + B = 90^{\circ} + k \cdot 180^{\circ} atau A - B = 90^{\circ} + k \cdot 180^{\circ} tidak berlaku langsung karena ada konstanta di dalam sinus dan kosinus.

Kita gunakan kembali \sin(x - 60^{\circ}) = \cos(x + 45^{\circ}).
Ubah \sin menjadi \cos atau sebaliknya.
\sin(x - 60^{\circ}) = \cos(90^{\circ} - (x - 60^{\circ})) = \cos(90^{\circ} - x + 60^{\circ}) = \cos(150^{\circ} - x).

Maka, \cos(150^{\circ} - x) = \cos(x + 45^{\circ}).
Ini berarti:
1. 150^{\circ} - x = x + 45^{\circ} + k \cdot 360^{\circ}
   105^{\circ} = 2x + k \cdot 360^{\circ}
   2x = 105^{\circ} - k \cdot 360^{\circ}
   x = 52.5^{\circ} - k \cdot 180^{\circ}
   Ini sama dengan x = 52.5^{\circ} + m \cdot 180^{\circ} (dengan m = -k).

2. 150^{\circ} - x = -(x + 45^{\circ}) + k \cdot 360^{\circ}
   150^{\circ} - x = -x - 45^{\circ} + k \cdot 360^{\circ}
   150^{\circ} = -45^{\circ} + k \cdot 360^{\circ}
   195^{\circ} = k \cdot 360^{\circ}
   Ini tidak mungkin untuk nilai k bulat.

Jadi, kita memiliki x = 52.5^{\circ}.
Untuk mencari \tan x = \tan 52.5^{\circ}:
Kita bisa gunakan identitas tan(A/2) = (1 - cos A) / sin A atau sin A / (1 + cos A).
Atau kita bisa gunakan \tan(A+B).
52.5^{\circ} = 22.5^{\circ} + 30^{\circ}
\tan 22.5^{\circ} = \sqrt{2} - 1
\tan 30^{\circ} = 1/\sqrt{3}

\tan(52.5^{\circ}) = \tan(22.5^{\circ} + 30^{\circ})
= \frac{\tan 22.5^{\circ} + \tan 30^{\circ}}{1 - \tan 22.5^{\circ} \tan 30^{\circ}}
= \frac{(\sqrt{2} - 1) + 1/\sqrt{3}}{1 - (\sqrt{2} - 1)(1/\sqrt{3})}
= \frac{\frac{\sqrt{6} - \sqrt{3} + 1}{\sqrt{3}}}{1 - \frac{\sqrt{2} - 1}{\sqrt{3}}}
= \frac{\frac{\sqrt{6} - \sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - \sqrt{2} + 1}{\sqrt{3}}}
= \frac{\sqrt{6} - \sqrt{3} + 1}{\sqrt{3} - \sqrt{2} + 1}

Mari kita coba cara lain: x = 52.5^{\circ} = (105/2)^{\circ}.
\tan(105^{\circ}) = \tan(60^{\circ} + 45^{\circ}) = \frac{\tan 60^{\circ} + \tan 45^{\circ}}{1 - \tan 60^{\circ} \tan 45^{\circ}} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{3 + 1 + 2\sqrt{3}}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}.

Sekarang gunakan \tan(A/2) = \frac{\sin A}{1 + \cos A}
\tan(52.5^{\circ}) = \tan(105^{\circ}/2) = \frac{\sin 105^{\circ}}{1 + \cos 105^{\circ}}
\sin 105^{\circ} = \sin(60^{\circ} + 45^{\circ}) = \sin 60^{\circ} \cos 45^{\circ} + \cos 60^{\circ} \sin 45^{\circ} = \frac{\sqrt{3}}{2} \frac{\sqrt{2}}{2} + \frac{1}{2} \frac{\sqrt{2}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}
\cos 105^{\circ} = \cos(60^{\circ} + 45^{\circ}) = \cos 60^{\circ} \cos 45^{\circ} - \sin 60^{\circ} \sin 45^{\circ} = \frac{1}{2} \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \frac{\sqrt{2}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4}

\tan(52.5^{\circ}) = \frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{1 + \frac{\sqrt{2} - \sqrt{6}}{4}} = \frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{4 + \sqrt{2} - \sqrt{6}}{4}} = \frac{\sqrt{6} + \sqrt{2}}{4 + \sqrt{2} - \sqrt{6}}
Kalikan dengan sekawan penyebut:
= \frac{(\sqrt{6} + \sqrt{2})(4 + \sqrt{6} - \sqrt{2})}{(4 + \sqrt{2} - \sqrt{6})(4 - \sqrt{2} + \sqrt{6})}
= \frac{4\sqrt{6} + 6 - \sqrt{12} + 4\sqrt{2} + \sqrt{12} - 2}{16 - (\sqrt{2} - \sqrt{6})^2}
= \frac{4\sqrt{6} + 4 + 4\sqrt{2}}{16 - (2 + 6 - 2\sqrt{12})}
= \frac{4\sqrt{6} + 4 + 4\sqrt{2}}{16 - (8 - 4\sqrt{3})}
= \frac{4\sqrt{6} + 4 + 4\sqrt{2}}{8 + 4\sqrt{3}}
= \frac{\sqrt{6} + 1 + \sqrt{2}}{2 + \sqrt{3}}
Kalikan lagi dengan sekawan:
= \frac{(\sqrt{6} + 1 + \sqrt{2})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}
= \frac{2\sqrt{6} - \sqrt{18} + 2 - \sqrt{3} + 2\sqrt{2} - \sqrt{6}}{4 - 3}
= \frac{\sqrt{6} - 3\sqrt{2} + 2 - \sqrt{3} + 2\sqrt{2}}{1}
= \sqrt{6} - \sqrt{2} + 2 - \sqrt{3}
Ini tidak terlihat benar.

Mari gunakan pendekatan lain:
\sin(x - 60) = \cos(x + 45)
\sin x \cos 60 - \cos x \sin 60 = \cos x \cos 45 - \sin x \sin 45
\sin x (1/2) - \cos x (\sqrt{3}/2) = \cos x (\sqrt{2}/2) - \sin x (\sqrt{2}/2)
Kalikan kedua sisi dengan 2:
\sin x - \sqrt{3} \cos x = \sqrt{2} \cos x - \sqrt{2} \sin x
Pindahkan semua suku \sin x ke kiri dan \cos x ke kanan:
\sin x + \sqrt{2} \sin x = \sqrt{3} \cos x + \sqrt{2} \cos x
\sin x (1 + \sqrt{2}) = \cos x (\sqrt{3} + \sqrt{2})
Bagi kedua sisi dengan \cos x dan (1 + \sqrt{2}):
\frac{\sin x}{\cos x} = \frac{\sqrt{3} + \sqrt{2}}{1 + \sqrt{2}}
\tan x = \frac{\sqrt{3} + \sqrt{2}}{1 + \sqrt{2}}
Sekarang rasionalkan penyebutnya:
\tan x = \frac{(\sqrt{3} + \sqrt{2})(1 - \sqrt{2})}{(1 + \sqrt{2})(1 - \sqrt{2})}
\tan x = \frac{\sqrt{3} - \sqrt{6} + \sqrt{2} - 2}{1 - 2}
\tan x = \frac{\sqrt{3} - \sqrt{6} + \sqrt{2} - 2}{-1}
\tan x = 2 - \sqrt{3} + \sqrt{6} - \sqrt{2}

Mari kita cek kembali hasil \tan(52.5^{\circ}).
\tan(52.5^{\circ}) = \tan(75^{\circ} - 22.5^{\circ}) = \frac{\tan 75^{\circ} - \tan 22.5^{\circ}}{1 + \tan 75^{\circ} \tan 22.5^{\circ}}
\tan 75^{\circ} = \tan(45^{\circ} + 30^{\circ}) = \frac{1 + \sqrt{3}}{\sqrt{3} - 1} = \frac{(1 + \sqrt{3})^2}{2} = \frac{1 + 3 + 2\sqrt{3}}{2} = 2 + \sqrt{3}
\tan 22.5^{\circ} = \sqrt{2} - 1
\tan(52.5^{\circ}) = \frac{(2 + \sqrt{3}) - (\sqrt{2} - 1)}{1 + (2 + \sqrt{3})(\sqrt{2} - 1)}
= \frac{3 + \sqrt{3} - \sqrt{2}}{1 + 2\sqrt{2} - 2 + \sqrt{6} - \sqrt{3}}
= \frac{3 + \sqrt{3} - \sqrt{2}}{-1 + 2\sqrt{2} + \sqrt{6} - \sqrt{3}}
Ini juga rumit.

Mari kita kembali ke \tan x = \frac{\sqrt{3} + \sqrt{2}}{1 + \sqrt{2}}
Untuk mendapatkan nilai numerik:
\sqrt{2} \approx 1.414
\sqrt{3} \approx 1.732
\tan x \approx \frac{1.732 + 1.414}{1 + 1.414} = \frac{3.146}{2.414} \approx 1.303
\arctan(1.303) \approx 52.5^{\circ}

Jadi, nilai \tan x adalah \frac{\sqrt{3} + \sqrt{2}}{1 + \sqrt{2}} atau bentuk rasionalnya 2 - \sqrt{3} + \sqrt{6} - \sqrt{2}.

Jawaban singkat: Nilai tan x adalah (√3 + √2) / (1 + √2).