Bài tự luận · Bài 18
Dạng 2. Rút gọn biểu thức lượng giác
Bài tự luận Công thức lượng giác · Bài 18
(CTST11) Rút gọn các biểu thức sau:
a) $\sin x\cos^{5}x-\cos x\sin^{5}x$
b) $\frac{\sin3x\cos2x+\sin x\cos6x}{\sin4x}$;
c) $\frac{\cos x-\cos2x+\cos3x}{\sin x-\sin2x+\sin3x}$
d) $\frac{2\sin(x+y)}{\cos(x+y)+\cos(x-y)}-\tan y$.
a) $\sin x\cos^{5}x-\cos x\sin^{5}x$
b) $\frac{\sin3x\cos2x+\sin x\cos6x}{\sin4x}$;
c) $\frac{\cos x-\cos2x+\cos3x}{\sin x-\sin2x+\sin3x}$
d) $\frac{2\sin(x+y)}{\cos(x+y)+\cos(x-y)}-\tan y$.
Xem lời giải
Lời giải
a) $\sin x\cos^{5}x-\cos x\sin^{5}x=\sin x\cos x\left( \cos^{4}x-\sin^{4}x \right)$
$\begin{array}{l} =\frac{1}{2}\sin2x\left( \cos^{2}x-\sin^{2}x \right)\left( \cos^{2}x+\sin^{2}x \right) \\ =\frac{1}{2}\sin2x\cos2x=\frac{1}{4}\sin4x \end{array}$
b) $\frac{\sin3x\cos2x+\sin x\cos6x}{\sin4x}=\frac{\frac{1}{2}(\sin x+\sin5x)+\frac{1}{2}[\sin(-5x)+\sin7x]}{\sin4x}$
$\begin{array}{l} =\frac{\sin x+\sin5x-\sin5x+\sin7x}{2\sin4x}=\frac{\sin x+\sin7x}{2\sin4x} \\ =\frac{2\sin4x\cos3x}{2\sin4x}=\cos3x; \end{array}$
c) $\frac{\cos x-\cos2x+\cos3x}{\sin x-\sin2x+\sin3x}=\frac{(\cos x+\cos3x)-\cos2x}{(\sin x+\sin3x)-\sin2x}=\frac{2\cos2x\cos x-\cos2x}{2\sin2x\cos x-\sin2x}$
$=\frac{\cos2x(2\cos x-1)}{\sin2x(2\cos x-1)}=\frac{\cos2x}{\sin2x}=\cot2x;$
d) $\frac{2\sin(x+y)}{\cos(x+y)+\cos(x-y)}-\tan y=\frac{2(\sin x\cos y+\cos x\sin y)}{2\cos x\cos y}-\tan y$
$\begin{array}{l} =\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}-\tan y=\tan x+\tan y-\tan y \\ =\tan x. \end{array}$
$\begin{array}{l} =\frac{1}{2}\sin2x\left( \cos^{2}x-\sin^{2}x \right)\left( \cos^{2}x+\sin^{2}x \right) \\ =\frac{1}{2}\sin2x\cos2x=\frac{1}{4}\sin4x \end{array}$
b) $\frac{\sin3x\cos2x+\sin x\cos6x}{\sin4x}=\frac{\frac{1}{2}(\sin x+\sin5x)+\frac{1}{2}[\sin(-5x)+\sin7x]}{\sin4x}$
$\begin{array}{l} =\frac{\sin x+\sin5x-\sin5x+\sin7x}{2\sin4x}=\frac{\sin x+\sin7x}{2\sin4x} \\ =\frac{2\sin4x\cos3x}{2\sin4x}=\cos3x; \end{array}$
c) $\frac{\cos x-\cos2x+\cos3x}{\sin x-\sin2x+\sin3x}=\frac{(\cos x+\cos3x)-\cos2x}{(\sin x+\sin3x)-\sin2x}=\frac{2\cos2x\cos x-\cos2x}{2\sin2x\cos x-\sin2x}$
$=\frac{\cos2x(2\cos x-1)}{\sin2x(2\cos x-1)}=\frac{\cos2x}{\sin2x}=\cot2x;$
d) $\frac{2\sin(x+y)}{\cos(x+y)+\cos(x-y)}-\tan y=\frac{2(\sin x\cos y+\cos x\sin y)}{2\cos x\cos y}-\tan y$
$\begin{array}{l} =\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}-\tan y=\tan x+\tan y-\tan y \\ =\tan x. \end{array}$