Bài 2 trang 15 Toán 11 Tập 1 | Cánh diều Giải Toán 11

Giải Toán 11 Bài 1: Góc lượng giác. Giá trị lượng giác của góc lượng giác

Bài 2 trang 15 Toán 11 Tập 1

Lời giải:

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

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

$\begin{array}{l}\cos \left(-{1035}^{\circ }\right)=\cos \left({1035}^{\circ }\right)=\cos \left({6.360}^{\circ }-{45}^{\circ }\right)\\ =\cos \left(-{45}^{\circ }\right)=\cos \left({45}^{\circ }\right)=\frac{\sqrt{2}}{2}\\ \sin \left(-{1035}^{\circ }\right)=-\sin \left({1035}^{\circ }\right)=-\sin \left({6.360}^{\circ }-{45}^{\circ }\right)\\ =-\sin \left(-{45}^{\circ }\right)=\sin \left({45}^{\circ }\right)=\frac{\sqrt{2}}{2}\\ \tan \left(-{1035}^{\circ }\right)=\frac{\sin \left(-{1035}^{\circ }\right)}{\cos \left(-{1035}^{\circ }\right)}=1\\ \cot \left(-{1035}^{\circ }\right)=\frac{1}{\tan \left(-{1035}^{\circ }\right)}=-1\end{array}$

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

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

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