penjelasan ada di gambar ya yang a sama b
TRigonoMetrI Dasar
Perbandingan
nilai fungsi
Penjelasan dengan langkah-langkah:
[tex]\sf a. \ \sin 60 + \tan 30 =[/tex]
[tex]\sf = \frac{1}{2}\sqrt 3 + \frac{1}{3}\sqrt 3[/tex]
[tex]\sf = \frac{5}{6}\sqrt 3\\[/tex]
[tex]\sf b. \dfrac{\sin150 + \sin 120}{\cos 120 - \cos 300}[/tex]
[tex]\sf =. \dfrac{\sin(180-30) + \sin (180-60)}{\cos (180-60) - \cos (360-60)}[/tex]
[tex]\sf=\dfrac{\frac{1}{2} + \frac{1}{2}\sqrt 3}{-\frac{1}{2} - \frac{1}{2}}[/tex]
[tex]\sf=\dfrac{\frac{1}{2} (1+\sqrt 3)}{-1} = -\frac{1}{2}(1 + \sqrt3)[/tex]
c. sin α = ¹/₃, α tumpul , maka cos α < 0
cos² α = 1 - sin² α
cos² α = 1 - (¹/₃)² = 1- ¹/₉ = ⁸/₉
cos α = ± [tex]\sf \frac{2}{3}\sqrt 2[/tex]
cos α < 0, maka cos α = - [tex]\sf \frac{2}{3}\sqrt 2[/tex]
d. tan β = p dan β di kuadran III
[tex]\sf misal \ \tan \beta =p \to \frac{d}{s} = \frac{p}{1}[/tex]
d=p , s= 1
m² = d² + s² = p² + 1
[tex]\sf m = \sqrt{p^2 + 1}[/tex]
[tex]\sf di\ KD \ III , nilai \ \sin \beta < 0[/tex]
[tex]\sf \sin \beta =- \frac{d}{m} =- \frac{p}{\sqrt{p^2 +1}}\ atau \\[/tex]
[tex]\sf \sin \beta = - \frac{p\sqrt{p^2 +1}}{p^2 + 1}[/tex]
