Rút gọn biểu thức $B=\frac{ an\left( \alpha-\frac{\pi}{2} ight)\cos\left( \frac{3\pi}{2}+\alpha ight)-\sin^{3}\left( \frac{7\pi}{2}-\alpha ight)}{\cos\left( \alpha-\frac{\pi}{2} ight) an\left( \frac{3\pi}{2}+\alpha ight)}$

Question

Rút gọn biểu thức $B=\frac{	an\left( \alpha-\frac{\pi}{2} ight)\cos\left( \frac{3\pi}{2}+\alpha ight)-\sin^{3}\left( \frac{7\pi}{2}-\alpha ight)}{\cos\left( \alpha-\frac{\pi}{2} ight)	an\left( \frac{3\pi}{2}+\alpha ight)}$

Beedemy · Accepted Answer

Lời giải: $\begin{array}{l} B=\frac{	an\left( \alpha-\frac{\pi}{2} ight)\cos\left( \frac{3\pi}{2}+\alpha ight)-\sin^{3}\left( \frac{7\pi}{2}-\alpha ight)}{\cos\left( \alpha-\frac{\pi}{2} ight)	an\left( \frac{3\pi}{2}+\alpha ight)}=\frac{-	an\left( \frac{\pi}{2}-\alpha ight).\cos\left( 2\pi-\frac{\pi}{2}+\alpha ight)-\sin^{3}\left( 3\pi+\frac{\pi}{2}-\alpha ight)}{\cos\left( \frac{\pi}{2}-\alpha ight).	an\left( 2\pi-\frac{\pi}{2}+\alpha ight)} \ \,\,\,=\frac{-\cot\alpha.\sin\alpha+\cos^{3}\alpha}{\sin\alpha.\left( -\cot\alpha ight)}=\frac{-\cos\alpha+\cos^{3}\alpha}{-\cos\alpha}=1-\cos^{2}\alpha=\sin^{2}\alpha. \end{array}$

Bài tập nâng cao Giá trị lượng giác của góc lượng giác · Bài 6