Bài tự luận · Bài 17
Dạng 2. Rút gọn biểu thức lượng giác
Bài tự luận Công thức lượng giác · Bài 17
(CTST11) Rút gọn các biểu thức sau:
a) $\frac{\sin\alpha+\sin2\alpha}{1+\cos\alpha+\cos2\alpha}$
b) $\frac{\cos\left( \frac{\pi}{4}+\alpha \right)-\cos\left( \frac{\pi}{4}-\alpha \right)}{\sin\left( \frac{\pi}{4}+\alpha \right)-\sin\left( \frac{\pi}{4}-\alpha \right)}$;
c) $\frac{\sin^{2}\alpha}{4-4\sin^{2}\frac{\alpha}{2}}$
a) $\frac{\sin\alpha+\sin2\alpha}{1+\cos\alpha+\cos2\alpha}$
b) $\frac{\cos\left( \frac{\pi}{4}+\alpha \right)-\cos\left( \frac{\pi}{4}-\alpha \right)}{\sin\left( \frac{\pi}{4}+\alpha \right)-\sin\left( \frac{\pi}{4}-\alpha \right)}$;
c) $\frac{\sin^{2}\alpha}{4-4\sin^{2}\frac{\alpha}{2}}$
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Lời giải
a) $\frac{\sin\alpha+\sin2\alpha}{1+\cos\alpha+\cos2\alpha}=\frac{\sin\alpha+2\sin\alpha\cos\alpha}{1+\cos\alpha+2\cos^{2}\alpha-1}=\frac{\sin\alpha(1+2\cos\alpha)}{\cos\alpha(1+2\cos\alpha)}=\frac{\sin\alpha}{\cos\alpha}=\tan\alpha$;
b) $\frac{\cos\left( \frac{\pi}{4}+\alpha \right)-\cos\left( \frac{\pi}{4}-\alpha \right)}{\sin\left( \frac{\pi}{4}+\alpha \right)-\sin\left( \frac{\pi}{4}-\alpha \right)}=\frac{-2\sin\frac{\frac{\pi}{2}}{2}\sin\frac{2\alpha}{2}}{2\cos\frac{\frac{\pi}{2}}{2}\sin\frac{2\alpha}{2}}=-\frac{\sin\frac{\pi}{4}\sin\alpha}{\cos\frac{\pi}{4}\sin\alpha}=-\tan\frac{\pi}{4}=-1$
c) $\frac{\sin^{2}\alpha}{4-4\sin^{2}\frac{\alpha}{2}}=\frac{\left( 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2} \right)^{2}}{4\left( 1-\sin^{2}\frac{\alpha}{2} \right)}=\frac{4\sin^{2}\frac{\alpha}{2}\cos^{2}\frac{\alpha}{2}}{4\cos^{2}\frac{\alpha}{2}}=\sin^{2}\frac{\alpha}{2}$
b) $\frac{\cos\left( \frac{\pi}{4}+\alpha \right)-\cos\left( \frac{\pi}{4}-\alpha \right)}{\sin\left( \frac{\pi}{4}+\alpha \right)-\sin\left( \frac{\pi}{4}-\alpha \right)}=\frac{-2\sin\frac{\frac{\pi}{2}}{2}\sin\frac{2\alpha}{2}}{2\cos\frac{\frac{\pi}{2}}{2}\sin\frac{2\alpha}{2}}=-\frac{\sin\frac{\pi}{4}\sin\alpha}{\cos\frac{\pi}{4}\sin\alpha}=-\tan\frac{\pi}{4}=-1$
c) $\frac{\sin^{2}\alpha}{4-4\sin^{2}\frac{\alpha}{2}}=\frac{\left( 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2} \right)^{2}}{4\left( 1-\sin^{2}\frac{\alpha}{2} \right)}=\frac{4\sin^{2}\frac{\alpha}{2}\cos^{2}\frac{\alpha}{2}}{4\cos^{2}\frac{\alpha}{2}}=\sin^{2}\frac{\alpha}{2}$