Diketahui nilai sin 15=p. Nilai tan 15 adalah...

p / sqrt(1-p^2)

Pembahasan

Untuk mencari nilai \(\tan 15\) jika diketahui \(\sin 15 = p\), kita dapat menggunakan identitas trigonometri.
Kita tahu bahwa \(\sin^2 \theta + \cos^2 \theta = 1
Jadi, \(\cos^2 15 = 1 - \sin^2 15
\(\cos^2 15 = 1 - p^2
\(\cos 15 = \sqrt{1 - p^2}
(Kita ambil akar positif karena sudut 15 derajat berada di kuadran I, di mana kosinus bernilai positif).

Sekarang, kita tahu bahwa \(\tan \theta = \frac{\sin \theta}{\cos \theta}
Jadi, \(\tan 15 = \frac{\sin 15}{\cos 15}
Substitusikan nilai \(\sin 15 = p\) dan \(\cos 15 = \sqrt{1 - p^2}\):
\(\tan 15 = \frac{p}{\sqrt{1 - p^2}}

Untuk memberikan nilai \(\tan 15\) yang lebih spesifik, kita bisa menghitung nilai \(\sin 15\) terlebih dahulu. Sudut 15 derajat dapat dipecah menjadi \(45 - 30\) atau \(60 - 45\).
Menggunakan \(\sin(A-B) = \sin A \cos B - \cos A \sin B\):
\(\sin 15 = \sin(45 - 30) = \sin 45 \cos 30 - \cos 45 \sin 30
\(\sin 15 = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{2}}{2})(\frac{1}{2})
\(\sin 15 = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}
Jadi, \(p = \frac{\sqrt{6} - \sqrt{2}}{4}

Sekarang kita cari \(\cos 15\) menggunakan \(\cos(A-B) = \cos A \cos B + \sin A \sin B\):
\(\cos 15 = \cos(45 - 30) = \cos 45 \cos 30 + \sin 45 \sin 30
\(\cos 15 = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})
\(\cos 15 = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}

Terakhir, \(\tan 15 = \frac{\sin 15}{\cos 15}
\(\tan 15 = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}}
\(\tan 15 = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}
Untuk merasionalkan penyebut, kalikan dengan konjugatnya:
\(\tan 15 = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} \times \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}
\(\tan 15 = \frac{(\sqrt{6} - \sqrt{2})^2}{(\sqrt{6})^2 - (\sqrt{2})^2}
\(\tan 15 = \frac{6 - 2\sqrt{12} + 2}{6 - 2}
\(\tan 15 = \frac{8 - 2(2\sqrt{3})}{4}
\(\tan 15 = \frac{8 - 4\sqrt{3}}{4}
\(\tan 15 = 2 - \sqrt{3}

Jadi, jika \(\sin 15 = p\), maka \(\tan 15 = \frac{p}{\sqrt{1-p^2}}\). Jika \(p = \frac{\sqrt{6} - \sqrt{2}}{4}\), maka \(\tan 15 = 2 - \sqrt{3}\).