Bài tập nâng cao · Bài 14
Dạng 4. Biến đổi tổng thành tích
Bài tập nâng cao Công thức lượng giác · Bài 14
Rút gọn các biểu thức sau:
a/ $A=\frac{\cos7x-\cos8x-\cos9x+\cos10x}{\sin7x-\sin8x-\sin9x+\sin10x}$
b/ $B=\frac{\sin2x+2\sin3x+\sin4x}{\sin3x+2\sin4x+\sin5x}$
c/ $C=\frac{1+\cos x+\cos2x+\cos3x}{\cos x+2\cos^{2}x-1}$
d/ $D=\frac{\sin4x+\sin5x+\sin6x}{\cos4x+\cos5x+\cos6x}$
a/ $A=\frac{\cos7x-\cos8x-\cos9x+\cos10x}{\sin7x-\sin8x-\sin9x+\sin10x}$
b/ $B=\frac{\sin2x+2\sin3x+\sin4x}{\sin3x+2\sin4x+\sin5x}$
c/ $C=\frac{1+\cos x+\cos2x+\cos3x}{\cos x+2\cos^{2}x-1}$
d/ $D=\frac{\sin4x+\sin5x+\sin6x}{\cos4x+\cos5x+\cos6x}$
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Lời giải
a/ $A=\frac{(\cos10x+\cos7x)-(\cos9x+\cos8x)}{(\sin10x+\sin7x)-(\sin9x+\sin8x)}=\frac{2\cos\frac{17x}{2}\cos\frac{3x}{2}-2\cos\frac{17x}{2}\cos\frac{x}{2}}{2\sin\frac{17x}{2}\cos\frac{3x}{2}-2\sin\frac{17x}{2}\cos\frac{x}{2}}$
$=\frac{2\cos\frac{17x}{2}(\cos\frac{3x}{2}-\cos\frac{x}{2})}{2\sin\frac{17x}{2}(\cos\frac{3x}{2}-\cos\frac{x}{2})}=\cot\frac{17x}{2}$
b/ $B=\frac{(\sin4x+\sin2x)+2\sin3x}{(\sin5x+\sin3x)+2\sin4x}=\frac{2\sin3x.\cos x+2\sin3x}{2\sin4x.\cos x+2\sin4x}$
$=\frac{2\sin3x(\cos x+1)}{2\sin4x(\cos x+1)}=\frac{\sin3x}{\sin4x}$
c/ $C=\frac{(\cos3x+\cos x)+1+2\cos^{2}x-1}{\cos x+(2\cos^{2}x-1)}=\frac{2\cos2x.\cos x+2\cos^{2}x}{\cos x+(2\cos^{2}x-1)}=\frac{2\cos x(\cos2x+\cos x)}{\cos x+\cos2x}=2\cos x$
d/ $D=\frac{(\sin6x+\sin4x)+\sin5x}{(\cos6x+\cos4x)+\cos5x}=\frac{2\sin5x.\cos x+\sin5x}{2\cos5x.\cos x+\cos5x}=\frac{\sin5x(2\cos x+1)}{\cos5x(2\cos x+1)}=\tan5x$
$=\frac{2\cos\frac{17x}{2}(\cos\frac{3x}{2}-\cos\frac{x}{2})}{2\sin\frac{17x}{2}(\cos\frac{3x}{2}-\cos\frac{x}{2})}=\cot\frac{17x}{2}$
b/ $B=\frac{(\sin4x+\sin2x)+2\sin3x}{(\sin5x+\sin3x)+2\sin4x}=\frac{2\sin3x.\cos x+2\sin3x}{2\sin4x.\cos x+2\sin4x}$
$=\frac{2\sin3x(\cos x+1)}{2\sin4x(\cos x+1)}=\frac{\sin3x}{\sin4x}$
c/ $C=\frac{(\cos3x+\cos x)+1+2\cos^{2}x-1}{\cos x+(2\cos^{2}x-1)}=\frac{2\cos2x.\cos x+2\cos^{2}x}{\cos x+(2\cos^{2}x-1)}=\frac{2\cos x(\cos2x+\cos x)}{\cos x+\cos2x}=2\cos x$
d/ $D=\frac{(\sin6x+\sin4x)+\sin5x}{(\cos6x+\cos4x)+\cos5x}=\frac{2\sin5x.\cos x+\sin5x}{2\cos5x.\cos x+\cos5x}=\frac{\sin5x(2\cos x+1)}{\cos5x(2\cos x+1)}=\tan5x$