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a)
\(\cos\dfrac{22\pi}{3}=\cos\left(8\pi-\dfrac{2\pi}{3}\right)\\ =\cos\left(-\dfrac{2\pi}{3}\right)\\ =\cos\left(\dfrac{2\pi}{3}\right)\\ =-\cos\dfrac{\pi}{3}\\ =-\dfrac{1}{2}\)
b)
\(\sin\dfrac{23\pi}{4}=\sin\left(6\pi-\dfrac{\pi}{4}\right)\\ =\sin\left(-\dfrac{\pi}{4}\right)\\ =-\dfrac{\sqrt{2}}{2}\)
c)
\(\sin\dfrac{25\pi}{3}-\tan\dfrac{10\pi}{3}\\ =\sin\left(8\pi+\dfrac{\pi}{3}\right)-\tan\left(3\pi+\dfrac{\pi}{3}\right)\\ =\sin\dfrac{\pi}{3}-\tan\dfrac{\pi}{3}\\ =\dfrac{\sqrt{3}}{2}-\sqrt{3}\\ =\dfrac{-\sqrt{3}}{2}\)
d)
\(\cos^2\dfrac{\pi}{8}-\sin^2\dfrac{\pi}{8}\\ =\cos\dfrac{\pi}{4}\\ =\dfrac{\sqrt{2}}{2}\)
cau a: \(cos\dfrac{22\Pi}{3}=cos\dfrac{24\Pi-2\Pi}{3}=cos\left(8\Pi-\dfrac{2\Pi}{3}\right)=cos\dfrac{2\Pi}{3}=-\dfrac{1}{2}\)
câu b: \(sin\dfrac{23\Pi}{4}=sin\dfrac{24\Pi-\Pi}{4}=sin\left(6\Pi-\dfrac{\Pi}{4}\right)=-sin\dfrac{\Pi}{4}=-\dfrac{\sqrt{2}}{2}\)
cau c: \(=sin\left(8\Pi-\dfrac{\Pi}{3}\right)-tan\left(3\Pi+\dfrac{\Pi}{3}\right)=-sin\dfrac{\Pi}{3}-tan\dfrac{\Pi}{3}=-\dfrac{\sqrt{3}}{2}-\sqrt{3}=\dfrac{-3\sqrt{3}}{2}\)
cau d: \(cos^2\dfrac{\Pi}{8}-sin^2\dfrac{\Pi}{8}=cos2\left(\dfrac{\Pi}{8}\right)=cos\dfrac{\Pi}{4}=\dfrac{\sqrt{2}}{2}\)
rút gọn biểu thức:
E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))
Lời giải:
Vì \(\frac{\pi}{2}< \alpha< \pi \Rightarrow \frac{\pi}{4}< \frac{\alpha}{2}< \frac{\pi}{2}\)
\(\Rightarrow 1> \sin \frac{\alpha}{2}, \cos \frac{\alpha}{2}>0(*)\)
Mà \(\frac{-5}{9}=\sin \alpha=2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}<0\), điều này hoàn toàn mâu thuẫn với $(*)$
Bạn xem lại đề bài.
a) Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha< 0;cot\alpha>0;tan\alpha>0\).
Vì vậy: \(sin\alpha=-\sqrt{1-cos^2\alpha}=\dfrac{-\sqrt{15}}{4}\).
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\sqrt{15}}{4}:\dfrac{-1}{4}=\sqrt{15}\).
\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{\sqrt{15}}\).
b) Do \(\dfrac{\pi}{2}< \alpha< \pi\) nên \(cos\alpha< 0;tan\alpha< 0;cot\alpha< 0\).
\(cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{\sqrt{5}}{3}\);
\(tan\alpha=\dfrac{2}{3}:\dfrac{-\sqrt{5}}{3}=\dfrac{-2}{\sqrt{5}}\); \(cot\alpha=1:tan\alpha=\dfrac{-\sqrt{5}}{2}\).
\(\dfrac{\pi}{2}< a< \pi\) => sina > 0, cosa < 0
cos2a = \(\pm\sqrt{1-sin^22a}=\pm\sqrt{1-\left(\dfrac{5}{9}\right)^2}=\pm\dfrac{2\sqrt{14}}{9}\)
Nếu cos2a thì \(\dfrac{2\sqrt{14}}{9}\) thì
sina \(=\sqrt{\dfrac{1-cos2a}{2}}=\sqrt{\dfrac{1-\dfrac{2\sqrt{14}}{9}}{2}}=\dfrac{\sqrt{9-2\sqrt{14}}}{3\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}}{3\sqrt{2}}=\dfrac{\sqrt{7}-\sqrt{2}}{3\sqrt{2}}=\dfrac{\sqrt{14}-2}{6}\)
Nếu cos2a \(=-\dfrac{2\sqrt{14}}{9}\)
thì sina \(=\sqrt{\dfrac{1cos2a}{2}}=\sqrt{\dfrac{1+\dfrac{2\sqrt{14}}{9}}{2}}=\dfrac{2\sqrt{14}}{6}\)
cosa \(=-\sqrt{\dfrac{1+cos2a}{2}}=-\sqrt{\dfrac{9-2\sqrt{14}}{18}}=\dfrac{2-\sqrt{14}}{6}\)