Bài tập 1.10 trang 10 SBT Toán 11 Tập 1 Kết nối tri thức - KNTT

Ta có:

\(\begin{array}{l}\cos {105^0} = \cos ({60^0} + {45^0}) = \cos 60{\,^0}\cos {45^0} - \sin {60^0}\sin {45^0}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}.\frac{{\sqrt 2 }}{2} - \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 - \sqrt 6 }}{4}.\\\sin {105^0} = \sin ({60^0} + {45^0}) = \sin 60{\,^0}\cos {45^0} - \cos {60^0}\sin {45^0}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}.\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 + \sqrt 6 }}{4}.\end{array}\)

\(\tan {105^0} = \frac{{\sin {{105}^0}}}{{\cos {{105}^0}}} = \frac{{\frac{{\sqrt 2 - \sqrt 6 }}{4}}}{{\frac{{\sqrt 2 + \sqrt 6 }}{4}}} = \frac{{\sqrt 2 - \sqrt 6 }}{{\sqrt 2 + \sqrt 6 }}\).

\(\cot {105^0} = \frac{1}{{\tan {{105}^0}}} = 1:\frac{{\sqrt 2 - \sqrt 6 }}{{\sqrt 2 + \sqrt 6 }} = \frac{{\sqrt 2 + \sqrt 6 }}{{\sqrt 2 - \sqrt 6 }}\).

-- Mod Toán 11 HỌC247