Matematika Sekolah Menengah Atas a. sin 15⁰b. cos 165⁰c. Tan 105⁰​

elenorasmithamulv – March 05, 2023

a. sin 15⁰
b. cos 165⁰
c. Tan 105⁰
​

Penjelasan dengan langkah-langkah:

a) sin 15°

[tex] ⟹ \sin(a - b) \\ ⟹ \sin(a) \cos(b) - \cos(a) \sin(b) \\ ⟹sin(45 - 30) \\ ⟹ \sin(45) \cos(30 )- \cos(45) \sin(30) \\ ⟹ \sin(15) = \frac{1}{ \sqrt{2} } - \frac{ \sqrt{3} }{2} - \frac{1}{ \sqrt{2} } . \frac{1}{2} \\ ⟹ \frac{ \sqrt{3} - 1}{2 \sqrt{2} } ...[/tex]

b) cos 165°

[tex]⟹ \cos(30 + 135) \\ ⟹ \cos(30) \cos(135) - \sin(30) \sin(135) \\ ⟹ ( \frac{ \sqrt{3} }{2} )( - \frac{ \sqrt{2} }{2} ) - ( \frac{1}{2} )( \frac{ \sqrt{2} }{2} ) \\ ⟹ - \frac{ \sqrt{6} }{4} - \frac{ \sqrt{2} }{4} \\ ⟹ - \frac{1}{4} ( \sqrt{6} + \sqrt{2} )...[/tex]

c) tan 105°

[tex]⟹ \tan(60 + 45) \\ ⟹ \frac{ \tan(60) + \tan(45) }{1 - \tan(60) . \tan(45) } \\ ⟹ \frac{ \sqrt{3} + 1 }{1 - \sqrt{3} } \\ ⟹ - \frac{1}{2} ( \sqrt{3} + 1) {}^{2} \\ ⟹ - (2 + \sqrt{3} )..[/tex]

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