Bài tập nâng cao · Bài 10
Dạng 4. Biến đổi tổng thành tích
Bài tập nâng cao Công thức lượng giác · Bài 10
Biến đổi thành tích
a, $A=2\sin4x+\sqrt{2}$
b, $B=3-4\cos^{2}x$
c, $D=\sin2x+\sin4x+\sin6x$
d, $E=3+4\cos4x+\cos8x$
e, $F=\sin5x+\sin6x+\sin7x+\sin8x$
f, $G=1+\sin2x-\cos2x-\tan2x$
g, $H=\sin^{2}(x+90^{\circ})-3\cos^{2}(x-90^{\circ})$
h, $L=1+\sin x+\cos x$
a, $A=2\sin4x+\sqrt{2}$
b, $B=3-4\cos^{2}x$
c, $D=\sin2x+\sin4x+\sin6x$
d, $E=3+4\cos4x+\cos8x$
e, $F=\sin5x+\sin6x+\sin7x+\sin8x$
f, $G=1+\sin2x-\cos2x-\tan2x$
g, $H=\sin^{2}(x+90^{\circ})-3\cos^{2}(x-90^{\circ})$
h, $L=1+\sin x+\cos x$
Xem lời giải
Lời giải
a, $A=2\left( \sin4x+\frac{\sqrt{2}}{2} \right)=2\left( \sin4x+\sin\frac{\pi}{4} \right)=4\sin\left( 2x+\frac{\pi}{8} \right)\cos\left( 2x-\frac{\pi}{8} \right)$.
b, $B=1-2\cos2x=2\left( \frac{1}{2}-\cos2x \right)=2\left( \cos\frac{\pi}{3}-\cos2x \right)=4\sin\left( x+\frac{\pi}{6} \right)\sin\left( x-\frac{\pi}{6} \right)$.
c, $D=\sin2x+\sin6x+\sin4x=2\sin4x\cos2x+\sin4x=\sin4x(2\cos2x+1)$
$=2\sin4x\left( \cos2x+\cos\frac{\pi}{3} \right)=4\sin4x\cos\left( x+\frac{\pi}{6} \right)\cos\left( x-\frac{\pi}{6} \right).$
d, $E=3+4\cos4x+2\cos^{2}4x-1=2(\cos4x+1)^{2}=8\cos^{4}2x.$
e, $F=\sin5x+\sin8x+\sin6x+\sin7x=2\sin\frac{13x}{2}\left( \cos\frac{3x}{2}+\cos\frac{x}{2} \right)=4\sin\frac{13x}{2}\cos x\cos\frac{x}{2}.$
f, $G=1-\cos2x+\sin2x-\tan2x=\left( 1-\cos2x \right)\left( 1-\tan2x \right)=2\sin^{2}x\left( \tan\frac{\pi}{4}-\tan2x \right)$
$=\frac{2\sin^{2}x\sin\left( -2x+\frac{\pi}{4} \right)}{\cos\frac{\pi}{4}.\cos2x}=\frac{2\sqrt{2}\sin^{2}x\sin\left( -2x+\frac{\pi}{4} \right)}{\cos2x}.$
g, $H=\cos^{2}x-3\sin^{2}x=\left( \cos x-\sqrt{3}\sin x \right)\left( \cos x+\sqrt{3}\sin x \right)$
$=4\left( \frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x \right)\left( \frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x \right)=4\cos\left( x+\frac{\pi}{3} \right)\cos\left( x-\frac{\pi}{3} \right).$
h, $L=\sin x+\left( \cos x+1 \right)=2\sin\frac{x}{2}\cos\frac{x}{2}+2\cos^{2}\frac{x}{2}=2\cos\frac{x}{2}\left( \cos\frac{x}{2}+\sin\frac{x}{2} \right)=2\sqrt{2}\cos\frac{x}{2}\sin\left( \frac{x}{2}+\frac{\pi}{4} \right).$
b, $B=1-2\cos2x=2\left( \frac{1}{2}-\cos2x \right)=2\left( \cos\frac{\pi}{3}-\cos2x \right)=4\sin\left( x+\frac{\pi}{6} \right)\sin\left( x-\frac{\pi}{6} \right)$.
c, $D=\sin2x+\sin6x+\sin4x=2\sin4x\cos2x+\sin4x=\sin4x(2\cos2x+1)$
$=2\sin4x\left( \cos2x+\cos\frac{\pi}{3} \right)=4\sin4x\cos\left( x+\frac{\pi}{6} \right)\cos\left( x-\frac{\pi}{6} \right).$
d, $E=3+4\cos4x+2\cos^{2}4x-1=2(\cos4x+1)^{2}=8\cos^{4}2x.$
e, $F=\sin5x+\sin8x+\sin6x+\sin7x=2\sin\frac{13x}{2}\left( \cos\frac{3x}{2}+\cos\frac{x}{2} \right)=4\sin\frac{13x}{2}\cos x\cos\frac{x}{2}.$
f, $G=1-\cos2x+\sin2x-\tan2x=\left( 1-\cos2x \right)\left( 1-\tan2x \right)=2\sin^{2}x\left( \tan\frac{\pi}{4}-\tan2x \right)$
$=\frac{2\sin^{2}x\sin\left( -2x+\frac{\pi}{4} \right)}{\cos\frac{\pi}{4}.\cos2x}=\frac{2\sqrt{2}\sin^{2}x\sin\left( -2x+\frac{\pi}{4} \right)}{\cos2x}.$
g, $H=\cos^{2}x-3\sin^{2}x=\left( \cos x-\sqrt{3}\sin x \right)\left( \cos x+\sqrt{3}\sin x \right)$
$=4\left( \frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x \right)\left( \frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x \right)=4\cos\left( x+\frac{\pi}{3} \right)\cos\left( x-\frac{\pi}{3} \right).$
h, $L=\sin x+\left( \cos x+1 \right)=2\sin\frac{x}{2}\cos\frac{x}{2}+2\cos^{2}\frac{x}{2}=2\cos\frac{x}{2}\left( \cos\frac{x}{2}+\sin\frac{x}{2} \right)=2\sqrt{2}\cos\frac{x}{2}\sin\left( \frac{x}{2}+\frac{\pi}{4} \right).$