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vậy thì kết quả là
\(\sin2\alpha=-0.96\)
\(\)còn \(\cos\left(\alpha+\frac{\pi}{6}\right)\) thì đúng vì -(-0.8) mà sorry thiếu ngủ hôm qua -_-
--.-- \(-\pi>-\frac{3}{2}\pi\) mà
Chắc nhầm đề rồi, phải là \(-\pi>a>-\frac{3}{2}\pi\)mới đúng chứ
\(-\pi>a>-\frac{3}{2}\pi\Leftrightarrow\pi>a>\frac{1}{2}\pi\)
\(\cos a=-\frac{4}{5}\Rightarrow\sin a=\frac{3}{5}\)
\(\sin2a=2\sin a.\cos a=2.\frac{3}{5}.\frac{-4}{5}=-\frac{24}{25}\)
\(\cos2a=2\cos^2a-1=\frac{7}{25}\)
\(\sin\left(\frac{5\pi}{2}-a\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a=-\frac{4}{5}\)
\(\sin\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{3}{5}-\frac{4}{5}.\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{10}\)
\(\cos\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{-4}{5}-\frac{\sqrt{2}}{2}.\frac{3}{5}=-\frac{7\sqrt{2}}{10}\)
\(\Rightarrow\tan\left(a+\frac{\pi}{4}\right)=\frac{1}{7}\)
\(\cos^2\left(\frac{a}{2}\right)=\frac{1+\cos a}{2}=\frac{1}{10}\Leftrightarrow\left|\cos\frac{a}{2}\right|=\frac{\sqrt{10}}{10}\)
Mà \(\frac{\pi}{2}>\frac{a}{2}>\frac{\pi}{4}\)
\(\Rightarrow\cos\frac{a}{2}=\frac{\sqrt{10}}{10}\)
a) \(\dfrac{tan2\alpha}{tan4\alpha-tan2\alpha}=\dfrac{sin2\alpha}{cos2\alpha}:\left(\dfrac{sin4\alpha}{cos4\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\right)\)
\(=\dfrac{sin2\alpha}{cos2\alpha}:\dfrac{sin4\alpha cos2\alpha-sin2\alpha cos4\alpha}{cos4\alpha cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos2\alpha}.\dfrac{cos4\alpha.cos2\alpha}{sin2\alpha}=cos4\alpha\).
b) \(\sqrt{1+sin\alpha}-\sqrt{1-sin\alpha}=\sqrt{sin^2\dfrac{\alpha}{2}+2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)\(-\sqrt{sin^2\dfrac{\alpha}{2}-2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)
\(=\sqrt{\left(sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right)^2}-\sqrt{\left(sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right)^2}\)
\(=\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
Vì \(0< \alpha< \dfrac{\pi}{2}\) nên \(0< \alpha< \dfrac{\pi}{4}\).
Trong \(\left(0;\dfrac{\pi}{4}\right)\) thì \(sin\dfrac{\alpha}{2}\) tăng dần từ 0 tới \(\dfrac{\sqrt{2}}{2}\) và \(cos\dfrac{\alpha}{2}\) giảm dần từ 1 tới \(\dfrac{\sqrt{2}}{2}\) nên \(\left|sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right|=-\left(sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right)=cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\).
Vì vậy:
\(\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
\(=sin\dfrac{\alpha}{4}+cos\dfrac{\alpha}{4}-\left(cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\right)=2sin\dfrac{\alpha}{4}\).
Bài 4:
$\sin a=\frac{1}{2}$ và $0< a< \pi$ nên $a=\frac{\pi}{6}$ hoặc $a=\frac{5}{6}\pi$
Nếu $a=\frac{\pi}{6}$ thì $\tan (2a-\frac{\pi}{2})+\sin a=\tan (2.\frac{\pi}{6}-\frac{\pi}{2})+\frac{1}{2}=\frac{-\sqrt{3}}{3}+\frac{1}{2}=\frac{3-2\sqrt{3}}{6}$
Nếu $a=\frac{5\pi}{6}$ thì:
\(\tan (2a-\frac{\pi}{2})+\sin a=\tan (2.\frac{5\pi}{6}-\frac{\pi}{2})+\frac{1}{2}=\frac{\sqrt{3}}{3}+\frac{1}{2}=\frac{3+2\sqrt{3}}{6}\)
Bài 3:
\(\tan a=\frac{-4}{7}=\frac{\sin a}{\cos a}\)
\(\Rightarrow \frac{\sin ^2a}{\cos ^2a}=\frac{16}{49}\Rightarrow \frac{1}{\cos ^2a}=\frac{65}{49}\) \(\Rightarrow \cos ^2a=\frac{49}{65}\)
Kết hợp điều kiện của $a$ suy ra $\cos a>0\Rightarrow \cos a=\frac{7}{\sqrt{65}}$
$\Rightarrow \sin a=\frac{-4}{7}\cos a=\frac{-4}{\sqrt{65}}$
Do đó:
\(\cos (2a-\frac{\pi}{2})=\cos 2a.\cos \frac{\pi}{2}+\sin 2a.\sin \frac{\pi}{2}\)
\(=(\cos ^2a-\sin ^2a).0+2\sin a\cos a.1=2\sin a\cos a=2.\frac{-4}{\sqrt{65}}.\frac{7}{\sqrt{65}}=\frac{56}{65}\)
\(VT=\frac{2\cos2\alpha.\cos\alpha}{2.\sin2\alpha\cos\alpha}.\frac{\sin2\alpha}{\cos2\alpha}-2\left(2\sin\alpha.\cos\alpha\right)^2\)
\(VT=1-2\left(\sin2\alpha\right)^2=\cos4\alpha\)
\(\frac{3\pi}{4}< a< \pi\Rightarrow\frac{3\pi}{2}< 2a< 2\pi\Rightarrow cos2a>0\)
\(\left(sina+cosa\right)^2=\frac{1}{4}\Leftrightarrow sin^2a+cos^2a+2sina.cosa=\frac{1}{4}\)
\(\Leftrightarrow1+sin2a=\frac{1}{4}\Rightarrow sin2a=-\frac{3}{4}\)
\(\Rightarrow cos2a=\sqrt{1-sin^22a}=\frac{\sqrt{7}}{4}\)
\(\Rightarrow tan2a=\frac{sin2a}{cos2a}=-\frac{3\sqrt{7}}{7}\)