anh cj giup e vs a
III. Câu trắc nghiệm trả lời ngắn
Câu 1: Cho góc \(\alpha\) thỏa mãn \(\frac{\pi}{2} < \alpha < \pi\) và \(\sin \alpha = \frac{4}{5}\). Tính \(P = \sin 2(\alpha + \pi)\).
Câu 2: Cho góc \(\alpha\) thỏa mãn \(0 < \alpha < \frac{\pi}{2}\) và \(\sin \alpha = \frac{2}{3}\). Tính \(P = \frac{1 + \sin 2\alpha + \cos 2\alpha}{\sin \alpha + \cos \alpha}\).
Câu 3: Biết \(\sin (\pi - \alpha) = -\frac{3}{5}\) và \(\pi < \alpha < \frac{3\pi}{2}\). Tính \(P = \sin \left( \alpha + \frac{\pi}{6} \right)\).
Câu 3:
\(\sin\left(\pi-\alpha\right)=-\frac35\)
=>\(\sin\alpha=-\frac35\)
\(\pi<\alpha<\frac32\pi\)
=>\(cos\alpha<0\)
Ta có: \(\sin^2\alpha+cos^2\alpha=1\)
=>\(cos^2\alpha=1-\left(-\frac35\right)^2=1-\frac{9}{25}=\frac{16}{25}\)
mà \(cos\alpha<0\)
nên \(cos\alpha=-\sqrt{\frac{16}{25}}=-\frac45\)
\(P=\sin\left(\alpha+\frac{\pi}{6}\right)\)
\(=\sin\left(\alpha\right)\cdot cos\left(\frac{\pi}{6}\right)+cos\left(\alpha\right)\cdot\sin\left(\frac{\pi}{6}\right)\)
\(=\frac{-3}{5}\cdot\frac{\sqrt3}{2}+\frac{-4}{5}\cdot\frac12=\frac{-3\sqrt3+4}{10}\)
Câu 2:
\(\sin^2\alpha+cos^2\alpha=1\)
=>\(cos^2\alpha=1-\left(\frac23\right)^2=1-\frac49=\frac59\)
mà \(cos\alpha>0\left(0<\alpha<\frac{\pi}{2}\right)\)
nên \(cos\alpha=\sqrt{\frac59}=\frac{\sqrt5}{3}\)
\(P=\frac{1+\sin2\alpha+cos2\alpha}{\sin\alpha+cos\alpha}\)
\(=\frac{1+2\cdot\sin\alpha\cdot cos\alpha+2\cdot cos^2\alpha-1}{\sin\alpha+cos\alpha}=\frac{2\cdot\sin\alpha\cdot cos\alpha+2\cdot cos^2\alpha}{\sin\alpha+cos\alpha}\)
\(=\frac{2\cdot cos\alpha\left(\sin\alpha+cos\alpha\right)}{\sin\alpha+cos\alpha}=2\cdot cos\alpha=\frac{2\sqrt5}{3}\)
Câu 1:
Ta có: \(\sin^2\alpha+cos^2\alpha=1\)
=>\(cos^2\alpha=1-\left(\frac45\right)^2=1-\frac{16}{25}=\frac{9}{25}\)
mà \(cos\alpha<0\left(\frac{\pi}{2}<\alpha<\pi\right)\)
nên \(cos\alpha=-\sqrt{\frac{9}{25}}=-\frac35\)
\(\sin\left\lbrack2\left(\alpha+\pi\right)\right\rbrack=\sin\left\lbrack2\alpha+2\pi\right\rbrack=\sin2\alpha\)
\(=2\cdot\sin\alpha\cdot cos\alpha\)
\(=2\cdot\frac45\cdot\frac{-3}{5}=\frac{-24}{25}\)
