1. Cho [TEX]\cos\alpha=\displaystyle \frac{-3}{4}[/TEX] và [TEX]-\pi<\alpha<\displaystyle \frac{-\pi}{4}[/TEX]. Tính [TEX]A=\displaystyle \frac{\displaystyle \frac{1}{2}\sin\alpha-\tan\alpha+\displaystyle \frac{2}{3\cos\alpha}}{\cos\alpha-2\sin\alpha-\displaystyle \frac{3}{5}\cos\alpha}[/TEX]
3. Hãy tính: [TEX]C=\displaystyle \frac{3\sin^2\displaystyle \frac{19\pi}{6}-2\cos^2\left(\displaystyle \frac{-8\pi}{3}\right)}{2\sin\displaystyle \frac{5\pi}{6}\cos\displaystyle \frac{5\pi}{3}}[/TEX]
HD thui nhé:
1. Chia làm hai trường hợp:
+ [TEX]-\pi<\alpha<\displaystyle \frac{-\pi}{2}[/TEX] sin < 0, tan > 0
Tính sin từ hệ thức: [tex]sin = -\sqrt{1-cos^{2}\alpha }[/tex]
Tính tan: [tex]tan\alpha = \sqrt{\frac{1}{cos^{2}\alpha } - 1}[/tex]
+ [TEX]\displaystyle \frac{-\pi}{2}<\alpha<\displaystyle \frac{-\pi}{4}[/TEX] sin < 0, tan < 0
[tex]tan\alpha = -\sqrt{\frac{1}{cos^{2}\alpha } - 1}[/tex]
thay vào mà tính
3. [tex]sin\frac{19\Pi }{6} = sin(3\Pi + \frac{\Pi }{6})[/tex]; [tex]cos\frac{-8\Pi }{3} = cos(\frac{\Pi }{3} - 3\Pi )[/tex]
[tex]sin\frac{5\Pi }{6} = sin(\Pi - \frac{\Pi }{6} )[/tex]; [tex]cos\frac{5\Pi }{3} = cos(\Pi + \frac{2\Pi }{3})[/tex]
