Penjelasan dengan langkah-langkah:
[tex] \cos(a) = \frac{4}{12} \to \frac{sa}{mi} [/tex]
[tex] \: [/tex]
—Karena sudut a merupakan sudut lancip, maka nilai sinus positif, nilai tangen positif, dan nilai cotangen positif. Sekarang cari de.
[tex]de = \sqrt{ {12}^{2} - {4}^{2} } [/tex]
[tex]de = \sqrt{144 - 16} [/tex]
[tex]de = \sqrt{128} [/tex]
[tex]de = 8 \sqrt{2} [/tex]
[tex] \: [/tex]
—Hitung operasi tersebut.
[tex] = \sin(a) + \tan(a) \cos(a) - \cot(a) [/tex]
[tex] = \sin(a) + \frac{ \sin(a) }{ \cancel{\cos(a)} }. \cancel{ \cos(a)} - \cot(a) [/tex]
[tex] = \sin(a) + \sin(a) - \cot(a) [/tex]
[tex] = 2 \sin(a) - \cot(a) [/tex]
[tex] = 2. \frac{de}{mi} - \frac{sa}{de} [/tex]
[tex] = 2. \frac{8 \sqrt{2} }{12} - \frac{4}{8 \sqrt{2} } [/tex]
[tex] = \cancel2. \frac{8 \sqrt{2} }{ \cancel{12}} - \frac{ 4}{8 \sqrt{2} } [/tex]
[tex] = 1. \frac{8 \sqrt{2} }{ 6} - \frac{ \cancel4}{ \cancel8 \sqrt{2} } [/tex]
[tex] = 1. \frac{ \cancel8 \sqrt{2} }{ \cancel6} - \frac{ 1}{ 2\sqrt{2} } [/tex]
[tex] = \frac{3 \sqrt{2} }{2} - \frac{1}{2 \sqrt{2} } [/tex]
[tex] = \frac{3 \sqrt{2} }{2} - \frac{2 \sqrt{2} }{4.2} [/tex]
[tex] = \frac{3 \sqrt{2} }{2} - \frac{ \sqrt{2} }{4} [/tex]
[tex] = \frac{6 \sqrt{2} }{4} - \frac{ \sqrt{2} }{4} [/tex]
[tex] = \frac{5 \sqrt{2} }{4} \: \: \: \text{jawabannya}[/tex]
[answer.2.content]
