Cho Sina = 3/5 , π/2 < a<π Tính cos( a + 2π/3 ) 30/08/2021 Bởi Melanie Cho Sina = 3/5 , π/2 < a<π Tính cos( a + 2π/3 )
$\dfrac{\pi}{2}<a\pi\to \cos a<0$ $\to \cos a=-\sqrt{1-\sin^2a}=-\sqrt{1-\Big(\dfrac{3}{5}\Big)^2}=\dfrac{-4}{5}$ $\cos\Big(a+\dfrac{2\pi}{3}\Big)$ $=\cos a\cos\dfrac{2\pi}{3}-\sin a\sin\dfrac{2\pi}{3}$ $=\dfrac{-4}{5}.\dfrac{-1}{2}-\dfrac{3}{5}.\dfrac{\sqrt3}{2}$ $=\dfrac{4-3\sqrt3}{10}$ Bình luận
Giải thích các bước giải: Ta có: $\sin a=\dfrac35$ $\to \cos^2a=1-\sin^2a=\dfrac{16}{25}$ Mà $\dfrac{\pi}{2}<a<\pi \to \cos a<0$ $\to \cos a=-\dfrac45$ Ta có: $\cos(a+\dfrac23\pi)=\cos a\cos(\dfrac23\pi)-\sin a\sin(\dfrac23\pi)$ $\to \cos(a+\dfrac23\pi)=-\dfrac45\cdot (-\dfrac12)-\dfrac35\cdot \dfrac{\sqrt{3}}{2}$ $\to \cos(a+\dfrac23\pi)=\dfrac{4-3\sqrt3}{10}$ Bình luận
$\dfrac{\pi}{2}<a\pi\to \cos a<0$
$\to \cos a=-\sqrt{1-\sin^2a}=-\sqrt{1-\Big(\dfrac{3}{5}\Big)^2}=\dfrac{-4}{5}$
$\cos\Big(a+\dfrac{2\pi}{3}\Big)$
$=\cos a\cos\dfrac{2\pi}{3}-\sin a\sin\dfrac{2\pi}{3}$
$=\dfrac{-4}{5}.\dfrac{-1}{2}-\dfrac{3}{5}.\dfrac{\sqrt3}{2}$
$=\dfrac{4-3\sqrt3}{10}$
Giải thích các bước giải:
Ta có:
$\sin a=\dfrac35$
$\to \cos^2a=1-\sin^2a=\dfrac{16}{25}$
Mà $\dfrac{\pi}{2}<a<\pi \to \cos a<0$
$\to \cos a=-\dfrac45$
Ta có:
$\cos(a+\dfrac23\pi)=\cos a\cos(\dfrac23\pi)-\sin a\sin(\dfrac23\pi)$
$\to \cos(a+\dfrac23\pi)=-\dfrac45\cdot (-\dfrac12)-\dfrac35\cdot \dfrac{\sqrt{3}}{2}$
$\to \cos(a+\dfrac23\pi)=\dfrac{4-3\sqrt3}{10}$