Bài tự luận · Bài 21
Dạng 3. Chứng minh đẳng thức lượng giác
Bài tự luận Công thức lượng giác · Bài 21
(CTST11) Chứng minh các đẳng thức lượng giác sau:
a) $4\cos x\cos\left( \frac{\pi}{3}-x \right)\cos\left( \frac{\pi}{3}+x \right)=\cos 3x$
b) $\frac{\sin 2x\cos x}{(1+\cos x)(1+\cos 2x)}=\tan\frac{x}{2}$;
c) $\sin x(1+2\cos 2x+2\cos 4x+2\cos 6x)=\sin 7x$;
d) $\frac{\sin^{2}3x}{\sin^{2}x}-\frac{\cos^{2}3x}{\cos^{2}x}=8\cos 2x$.
a) $4\cos x\cos\left( \frac{\pi}{3}-x \right)\cos\left( \frac{\pi}{3}+x \right)=\cos 3x$
b) $\frac{\sin 2x\cos x}{(1+\cos x)(1+\cos 2x)}=\tan\frac{x}{2}$;
c) $\sin x(1+2\cos 2x+2\cos 4x+2\cos 6x)=\sin 7x$;
d) $\frac{\sin^{2}3x}{\sin^{2}x}-\frac{\cos^{2}3x}{\cos^{2}x}=8\cos 2x$.
Xem lời giải
Lời giải
a) $4\cos x\cos\left( \frac{\pi}{3}-x \right)\cos\left( \frac{\pi}{3}+x \right)=2\cos x\left( \cos 2x+\cos\frac{2\pi}{3} \right)$
$\begin{array}{l}=2\cos x\cos 2x+2\cos x\cos\frac{2\pi}{3} \\ =\cos x+\cos 3x+2\cos x\cdot \left( -\frac{1}{2} \right) \\ =\cos x+\cos 3x-\cos x=\cos 3x\end{array}$
b)
$\begin{array}{ll} \frac{\sin 2x\cos x}{(1+\cos x)(1+\cos 2x)} & =\frac{(2\sin x\cos x)\cos x}{\left( 1+2\cos^{2}\frac{x}{2}-1 \right)\left( 1+2\cos^{2}x-1 \right)}=\frac{2\sin x\cos^{2}x}{4\cos^{2}\frac{x}{2}\cos^{2}x} \\ & =\frac{\sin x}{2\cos^{2}\frac{x}{2}}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^{2}\frac{x}{2}}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2}\end{array}$
c)
$\begin{array}{l}\sin x(1+2\cos 2x+2\cos 4x+2\cos 6x) \\ =\sin x+2\sin x\cos 2x+2\sin x\cos 4x+2\sin x\cos 6x \\ =\sin x+[\sin(-x)+\sin 3x]+[\sin(-3x)+\sin 5x]+[\sin(-5x)+\sin 7x] \\ =\sin x+(-\sin x+\sin 3x)+(-\sin 3x+\sin 5x)+(-\sin 5x+\sin 7x)=\sin 7x\end{array}$
d)
$\begin{array}{l}\frac{\sin^{2}3x}{\sin^{2}x}-\frac{\cos^{2}3x}{\cos^{2}x}=\frac{\sin^{2}3x\cos^{2}x-\cos^{2}3x\sin^{2}x}{\sin^{2}x\cos^{2}x}=\frac{(\sin 3x\cos x)^{2}-(\cos 3x\sin x)^{2}}{\sin^{2}x\cos^{2}x} \\ =\frac{(\sin 3x\cos x+\cos 3x\sin x)(\sin 3x\cos x-\cos 3x\sin x)}{\frac{1}{4}\sin^{2}2x} \\ =\frac{4\sin 4x\sin 2x}{\sin^{2}2x}=\frac{4(2\sin 2x\cos 2x)\sin 2x}{\sin^{2}2x} \\ =\frac{8\sin^{2}2x\cos 2x}{\sin^{2}2x}=8\cos 2x.\end{array}$
$\begin{array}{l}=2\cos x\cos 2x+2\cos x\cos\frac{2\pi}{3} \\ =\cos x+\cos 3x+2\cos x\cdot \left( -\frac{1}{2} \right) \\ =\cos x+\cos 3x-\cos x=\cos 3x\end{array}$
b)
$\begin{array}{ll} \frac{\sin 2x\cos x}{(1+\cos x)(1+\cos 2x)} & =\frac{(2\sin x\cos x)\cos x}{\left( 1+2\cos^{2}\frac{x}{2}-1 \right)\left( 1+2\cos^{2}x-1 \right)}=\frac{2\sin x\cos^{2}x}{4\cos^{2}\frac{x}{2}\cos^{2}x} \\ & =\frac{\sin x}{2\cos^{2}\frac{x}{2}}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^{2}\frac{x}{2}}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2}\end{array}$
c)
$\begin{array}{l}\sin x(1+2\cos 2x+2\cos 4x+2\cos 6x) \\ =\sin x+2\sin x\cos 2x+2\sin x\cos 4x+2\sin x\cos 6x \\ =\sin x+[\sin(-x)+\sin 3x]+[\sin(-3x)+\sin 5x]+[\sin(-5x)+\sin 7x] \\ =\sin x+(-\sin x+\sin 3x)+(-\sin 3x+\sin 5x)+(-\sin 5x+\sin 7x)=\sin 7x\end{array}$
d)
$\begin{array}{l}\frac{\sin^{2}3x}{\sin^{2}x}-\frac{\cos^{2}3x}{\cos^{2}x}=\frac{\sin^{2}3x\cos^{2}x-\cos^{2}3x\sin^{2}x}{\sin^{2}x\cos^{2}x}=\frac{(\sin 3x\cos x)^{2}-(\cos 3x\sin x)^{2}}{\sin^{2}x\cos^{2}x} \\ =\frac{(\sin 3x\cos x+\cos 3x\sin x)(\sin 3x\cos x-\cos 3x\sin x)}{\frac{1}{4}\sin^{2}2x} \\ =\frac{4\sin 4x\sin 2x}{\sin^{2}2x}=\frac{4(2\sin 2x\cos 2x)\sin 2x}{\sin^{2}2x} \\ =\frac{8\sin^{2}2x\cos 2x}{\sin^{2}2x}=8\cos 2x.\end{array}$