Cho sinα=45\sin\alpha = \dfrac{4}{5}sinα=54 và π2<α<π\dfrac{\pi}{2} < \alpha < \pi2π<α<π. Tính cosα\cos\alphacosα.
35\dfrac{3}{5}53.
−35- \dfrac{3}{5}−53.
−15- \dfrac{1}{5}−51.
15\dfrac{1}{5}51.
cos2α=1−sin2α=1−1625=925\cos^{2}\alpha = 1 - \sin^{2}\alpha = 1 - \dfrac{16}{25} = \dfrac{9}{25}cos2α=1−sin2α=1−2516=259.
Vì π2<α<π⇒cosα<0⇒cosα=−35\left. \dfrac{\pi}{2} < \alpha < \pi\Rightarrow\cos\alpha < 0\Rightarrow\cos\alpha = - \dfrac{3}{5} \right.2π<α<π⇒cosα<0⇒cosα=−53.
