The Collectors

Bài 6.36 trang 194 SBT đại số 10

Câu hỏi: Rút gọn các biểu thức
a) $\frac{\tan 2 \alpha}{\tan 4 \alpha-\tan 2 \alpha}$;
b) $\sqrt{1+\sin \alpha}-\sqrt{1-\sin \alpha},$ với $0<\alpha<\frac{\pi}{2}$;
c) $\frac{3-4 \cos 2 \alpha+\cos 4 \alpha}{3+4 \cos 2 \alpha+\cos 4 \alpha}$
d) $\frac{\sin \alpha+\sin 3 \alpha+\sin 5 \alpha}{\cos \alpha+\cos 3 \alpha+\cos 5 \alpha}$.
a) $\frac{\tan 2 \alpha}{\tan 4 \alpha-\tan 2 \alpha}=\frac{\tan 2 \alpha}{\frac{2 \tan 2 \alpha}{1}-\tan 2 \alpha}=\frac{1-\tan ^{2} 2 \alpha}{1+\tan ^{2} 2 \alpha}=\cos 4 \alpha$.
b) $\sqrt{1+\sin \alpha}-\sqrt{1-\sin \alpha}=\sqrt{\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)^{2}}-\sqrt{\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)^{2}}$
Vì $\quad 0<\alpha<\frac{\pi}{2}$ nên $0<\frac{\alpha}{2}<\frac{\pi}{4},$ suy ra $0<\sin \frac{\alpha}{2}<\cos \frac{\alpha}{2}$.
Vậy:
$
\begin{aligned}
\sqrt{1+\sin \alpha}-\sqrt{1-\sin \alpha} &=\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}-\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right) \\
&=2 \sin \frac{\alpha}{2}
\end{aligned}
$
c)
$\begin{aligned} \frac{3-4 \cos 2 \alpha+\cos 4 \alpha}{3+4 \cos 2 \alpha+\cos 4 \alpha} &=\frac{3-4 \cos 2 \alpha+2 \cos ^{2} 2 \alpha-1}{3+4 \cos 2 \alpha+2 \cos ^{2} 2 \alpha-1} \\ &=\frac{2\left(\cos ^{2} 2 \alpha-2 \cos 2 \alpha+1\right)}{2\left(\cos ^{2} 2 \alpha+2 \cos 2 \alpha+1\right)} \\ &=\frac{(\cos 2 \alpha-1)^{2}}{(\cos 2 \alpha+1)^{2}}=\frac{\left(-2 \sin ^{2} \alpha\right)^{2}}{\left(2 \cos ^{2} \alpha\right)^{2}}=\tan ^{4} \alpha \end{aligned}$
d)
$\begin{aligned} \frac{\sin \alpha+\sin 3 \alpha+\sin 5 \alpha}{\cos \alpha+\cos 3 \alpha+\cos 5 \alpha} &=\frac{(\sin 5 \alpha+\sin \alpha)+\sin 3 \alpha}{(\cos 5 \alpha+\cos \alpha)+\cos 3 \alpha} \\ &=\frac{\sin 3 \alpha(2 \cos 2 \alpha+1)}{\cos 3 \alpha(2 \cos 2 \alpha+1)}=\tan 3 \alpha \end{aligned}$
 

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