Cho cos ∝= $\frac{1}{3}$ và $270^{0}$< ∝< $360^{0}$ Tính sin ∝, tan ∝, cot ∝, cos $\frac{∝}{2}$ ,sin $\frac{∝}{2}$
Cho cos ∝= $\frac{1}{3}$ và $270^{0}$< ∝< $360^{0}$ Tính sin ∝, tan ∝, cot ∝, cos $\frac{∝}{2}$ ,sin $\frac{∝}{2}$
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
270^\circ < \alpha < 360^\circ \Rightarrow \left\{ \begin{array}{l}
\sin \alpha < 0\\
\cos \alpha > 0
\end{array} \right.\\
\sin \alpha < 0 \Rightarrow \sin \alpha = – \sqrt {1 – {{\cos }^2}\alpha } = – \sqrt {1 – {{\left( {\dfrac{1}{3}} \right)}^2}} = – \dfrac{{2\sqrt 2 }}{3}\\
\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = – 2\sqrt 2 \\
\cot \alpha = \dfrac{{\cos \alpha }}{{\sin \alpha }} = – \dfrac{1}{{2\sqrt 2 }}\\
270^\circ < \alpha < 360^\circ \Rightarrow 135^\circ < \dfrac{\alpha }{2} < 180^\circ \Rightarrow \left\{ \begin{array}{l}
\sin \dfrac{\alpha }{2} > 0\\
\cos \dfrac{\alpha }{2} < 0
\end{array} \right.\\
\cos \alpha = 2co{s^2}\dfrac{\alpha }{2} – 1 \Rightarrow {\cos ^2}\dfrac{\alpha }{2} = \dfrac{{\cos \alpha + 1}}{2} = \dfrac{2}{3}\\
\cos \dfrac{\alpha }{2} < 0 \Rightarrow \cos \dfrac{\alpha }{2} = – \dfrac{{\sqrt 6 }}{3}\\
\sin \dfrac{\alpha }{2} > 0 \Rightarrow \sin \dfrac{\alpha }{2} = \sqrt {1 – {{\cos }^2}\dfrac{\alpha }{2}} = \dfrac{{\sqrt 3 }}{3}
\end{array}\)