Bài 2 trang 15

Đề bài

Tính các giá trị lượng giác của mỗi góc sau: $225^{\circ}; - 225^{\circ}; - 1035^{\circ}$ ; $\frac{5 π}{3}; \frac{19 π}{2}; - \frac{159 π}{4}$

Phương pháp giải - Xem chi tiết

Sử dụng các công thức biến đổi lượng giác sau:

Lời giải chi tiết

$\begin{array}{l} \cos (225^{\circ}) = \cos (180^{\circ} + 45^{\circ}) = - \cos (45^{\circ}) = - \frac{\sqrt{2}}{2} \\ \sin (225^{\circ}) = \sin (180^{\circ} + 45^{\circ}) = - \sin (45^{\circ}) = - \frac{\sqrt{2}}{2} \\ \tan (225^{\circ}) = \frac{\sin (225^{\circ})}{\cos (225^{\circ})} = 1 \\ \cot (225^{\circ}) = \frac{1}{\tan (225^{\circ})} = 1 \end{array}$

$\begin{array}{l} \cos (- 225^{\circ}) = \cos (225^{\circ}) = \cos (180^{\circ} + 45^{\circ}) = - \cos (45^{\circ}) = - \frac{\sqrt{2}}{2} \\ \sin (- 225^{\circ}) = - \sin (225^{\circ}) = - \sin (180^{\circ} + 45^{\circ}) = \sin (45^{\circ}) = \frac{\sqrt{2}}{2} \\ \tan (- 225^{\circ}) = \frac{\sin (225^{\circ})}{\cos (225^{\circ})} = - 1 \\ \cot (- 225^{\circ}) = \frac{1}{\tan (225^{\circ})} = - 1 \end{array}$

$\begin{array}{l} \cos (- 1035^{\circ}) = \cos ({- 3.360}^{\circ} + 45^{\circ}) = \cos (45^{\circ}) = \frac{\sqrt{2}}{2} \\ \sin (- 1035^{\circ}) = \sin ({- 3.360}^{\circ} + 45^{\circ}) = \sin (45^{\circ}) = \frac{\sqrt{2}}{2} \\ \tan (- 1035^{\circ}) = \frac{\sin (- 1035^{\circ})}{\cos (- 1035^{\circ})} = 1 \\ \cot (- 1035^{\circ}) = \frac{1}{\tan (- 1035^{\circ})} = 1 \end{array}$

$\begin{array}{l} \cos (\frac{5 π}{3}) = \cos (π + \frac{2 π}{3}) = - \cos (\frac{2 π}{3}) = \frac{1}{2} \\ \sin (\frac{5 π}{3}) = \sin (π + \frac{2 π}{3}) = - \sin (\frac{2 π}{3}) = - \frac{\sqrt{3}}{2} \\ \tan (\frac{5 π}{3}) = \frac{\sin (\frac{5 π}{3})}{\cos (\frac{5 π}{3})} = - \sqrt{3} \\ \cot (\frac{5 π}{3}) = \frac{1}{\tan (\frac{5 π}{3})} = - \frac{\sqrt{3}}{3} \end{array}$

$\begin{array}{l} \cos (\frac{19 π}{2}) = \cos (8 π + \frac{3 π}{2}) = \cos (\frac{3 π}{2}) = \cos (π + \frac{π}{2}) = - \cos (\frac{π}{2}) = 0 \\ \sin (\frac{19 π}{2}) = \sin (8 π + \frac{3 π}{2}) = \sin (\frac{3 π}{2}) = \sin (π + \frac{π}{2}) = - \sin (\frac{π}{2}) = - 1 \\ \tan (\frac{19 π}{2}) \\ \cot (\frac{19 π}{2}) = \frac{\cos (\frac{19 π}{2})}{\sin (\frac{19 π}{2})} = 0 \end{array}$

$\begin{array}{l} \cos (- \frac{159 π}{4}) = \cos (\frac{159 π}{4}) = \cos (40. π - \frac{π}{4}) = \cos (- \frac{π}{4}) = \cos (\frac{π}{4}) = \frac{\sqrt{2}}{2} \\ \sin (- \frac{159 π}{4}) = - \sin (\frac{159 π}{4}) = - \sin (40. π - \frac{π}{4}) = - \sin (- \frac{π}{4}) = \sin (\frac{π}{4}) = \frac{\sqrt{2}}{2} \\ \tan (- \frac{159 π}{4}) = \frac{\cos (- \frac{159 π}{4})}{\sin (- \frac{159 π}{4})} = 1 \\ \cot (- \frac{159 π}{4}) = \frac{1}{\tan (- \frac{159 π}{4})} = 1 \end{array}$