Bài tự luận · Bài 20
Dạng 3. Chứng minh đẳng thức lượng giác
Bài tự luận Công thức lượng giác · Bài 20
(CTST11) Chứng minh các đẳng thức lượng giác sau:
a) $\sin\left( 60^{\circ}+\alpha \right)-\sin\left( 60^{\circ}-\alpha \right)=\sin\alpha$;
b) $\sin^{4}\alpha+\cos^{4}\alpha=\frac{3}{4}+\frac{1}{4}\cos4\alpha$;
c) $\sin\alpha(2\cos4\alpha+2\cos2\alpha+1)=\sin5\alpha$;
d) $\frac{\cos(\alpha-\beta)}{\cos(\alpha+\beta)}=\frac{1+\tan\alpha\tan\beta}{1-\tan\alpha\tan\beta}$.
a) $\sin\left( 60^{\circ}+\alpha \right)-\sin\left( 60^{\circ}-\alpha \right)=\sin\alpha$;
b) $\sin^{4}\alpha+\cos^{4}\alpha=\frac{3}{4}+\frac{1}{4}\cos4\alpha$;
c) $\sin\alpha(2\cos4\alpha+2\cos2\alpha+1)=\sin5\alpha$;
d) $\frac{\cos(\alpha-\beta)}{\cos(\alpha+\beta)}=\frac{1+\tan\alpha\tan\beta}{1-\tan\alpha\tan\beta}$.
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Lời giải
a) $\begin{array}{l} \sin\left( 60^{\circ}+\alpha \right)-\sin\left( 60^{\circ}-\alpha \right) \\ =\left( \sin60^{\circ}\cos\alpha+\cos60^{\circ}\sin\alpha \right)-\left( \sin60^{\circ}\cos\alpha-\cos60^{\circ}\sin\alpha \right) \\ =2\cos60^{\circ}\sin\alpha=2\cdot \frac{1}{2}\sin\alpha=\sin\alpha \end{array}$
b) $\begin{array}{l} \sin^{4}\alpha+\cos^{4}\alpha=\left( \sin^{2}\alpha+\cos^{2}\alpha \right)^{2}-2\sin^{2}\alpha\cos^{2}\alpha \\ =1-\frac{1}{2}\sin^{2}2\alpha=1-\frac{1}{4}(1-\cos4\alpha)=1-\frac{1}{4}+\frac{1}{4}\cos4\alpha \\ =\frac{3}{4}+\frac{1}{4}\cos4\alpha \end{array}$
c) $\begin{array}{l} \sin\alpha(2\cos4\alpha+2\cos2\alpha+1) \\ =2\sin\alpha\cos4\alpha+2\sin\alpha\cos2\alpha+\sin\alpha \\ =2\cdot \frac{1}{2}[\sin(-3\alpha)+\sin5\alpha]+2\cdot \frac{1}{2}[\sin(-\alpha)+\sin3\alpha]+\sin\alpha \\ =(-\sin3\alpha+\sin5\alpha)+(-\sin\alpha+\sin3\alpha)+\sin\alpha=\sin5\alpha \end{array}$
d) $\frac{\cos(\alpha-\beta)}{\cos(\alpha+\beta)}=\frac{\cos\alpha\cos\beta+\sin\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}=\frac{\cos\alpha\cos\beta(1+\tan\alpha\tan\beta)}{\cos\alpha\cos\beta(1-\tan\alpha\tan\beta)}=\frac{1+\tan\alpha\tan\beta}{1-\tan\alpha\tan\beta}.$
b) $\begin{array}{l} \sin^{4}\alpha+\cos^{4}\alpha=\left( \sin^{2}\alpha+\cos^{2}\alpha \right)^{2}-2\sin^{2}\alpha\cos^{2}\alpha \\ =1-\frac{1}{2}\sin^{2}2\alpha=1-\frac{1}{4}(1-\cos4\alpha)=1-\frac{1}{4}+\frac{1}{4}\cos4\alpha \\ =\frac{3}{4}+\frac{1}{4}\cos4\alpha \end{array}$
c) $\begin{array}{l} \sin\alpha(2\cos4\alpha+2\cos2\alpha+1) \\ =2\sin\alpha\cos4\alpha+2\sin\alpha\cos2\alpha+\sin\alpha \\ =2\cdot \frac{1}{2}[\sin(-3\alpha)+\sin5\alpha]+2\cdot \frac{1}{2}[\sin(-\alpha)+\sin3\alpha]+\sin\alpha \\ =(-\sin3\alpha+\sin5\alpha)+(-\sin\alpha+\sin3\alpha)+\sin\alpha=\sin5\alpha \end{array}$
d) $\frac{\cos(\alpha-\beta)}{\cos(\alpha+\beta)}=\frac{\cos\alpha\cos\beta+\sin\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}=\frac{\cos\alpha\cos\beta(1+\tan\alpha\tan\beta)}{\cos\alpha\cos\beta(1-\tan\alpha\tan\beta)}=\frac{1+\tan\alpha\tan\beta}{1-\tan\alpha\tan\beta}.$