Bài tập nâng cao · Bài 9
Dạng 3. Biến đổi tích thành tổng
Bài tập nâng cao Công thức lượng giác · Bài 9
Biến đổi thành tổng
a) $2\sin(a+b)\cos(a-b)$
b) $2\cos(a+b)\cos(a-b)$
c) $4\sin3x\sin2x\cos x$
d) $4\sin\frac{13x}{2}\cos x\cos\frac{x}{2}$
e) $\sin(x+30^{\circ})\cos(x-30^{\circ})$
f) $\sin\frac{\pi}{5}\sin\frac{2\pi}{5}$
g) $2\sin x\sin2x\sin3x$
h) $8\cos x\sin2x\sin3x$
i) $\sin\left( x+\frac{\pi}{6} \right)\sin\left( x-\frac{\pi}{6} \right)\cos2x$
k) $4\cos(a-b)\cos(b-c)\cos(c-a)$
a) $2\sin(a+b)\cos(a-b)$
b) $2\cos(a+b)\cos(a-b)$
c) $4\sin3x\sin2x\cos x$
d) $4\sin\frac{13x}{2}\cos x\cos\frac{x}{2}$
e) $\sin(x+30^{\circ})\cos(x-30^{\circ})$
f) $\sin\frac{\pi}{5}\sin\frac{2\pi}{5}$
g) $2\sin x\sin2x\sin3x$
h) $8\cos x\sin2x\sin3x$
i) $\sin\left( x+\frac{\pi}{6} \right)\sin\left( x-\frac{\pi}{6} \right)\cos2x$
k) $4\cos(a-b)\cos(b-c)\cos(c-a)$
Xem lời giải
Lời giải
a) $2\sin(a+b)\cos(a-b)=\sin2a+\sin2b$
b) $2\cos(a+b)\cos(a-b)=\cos2a+\cos2b$
c) $4\sin3x\sin2x\cos x=2\sin3x(\sin3x+\sin x)=2\sin^{2}3x+2\sin3x\sin x=2\sin^{2}3x-\cos4x+\cos2x$
d) $4\sin\frac{13x}{2}\cos x\cos\frac{x}{2}=2\sin\frac{13x}{2}(\cos\frac{3x}{2}+\cos\frac{x}{2})=2\sin\frac{13x}{2}\cos\frac{3x}{2}+2\sin\frac{13x}{2}\cos\frac{x}{2}=$
$=\sin8x+\sin5x+\sin7x+\sin6x$.
e) $\sin(x+30^{\circ})\cos(x-30^{\circ})=\frac{1}{2}(\sin2x+\sin60^{\circ})=\frac{1}{2}\sin2x+\frac{\sqrt{3}}{4}$
f) $\sin\frac{\pi}{5}\sin\frac{2\pi}{5}=-\frac{1}{2}(\cos\frac{3\pi}{5}-\cos\frac{\pi}{5})$
g) $2\sin x\sin2x\sin3x=\sin3x(\cos x-\cos3x)=\frac{1}{2}(2\sin3x\cos x-\sin6x)=\frac{1}{2}(\sin4x+\sin2x-\sin6x)$
h) $8\cos x\sin2x\sin3x=4\sin3x(\sin3x+\sin x)=4\sin^{2}3x+4\sin3x\sin x=4\sin^{2}3x+2\cos2x-2\cos4x$
i) $\sin\left( x+\frac{\pi}{6} \right)\sin\left( x-\frac{\pi}{6} \right)\cos2x=\frac{1}{2}\cos2x\left( \cos\frac{\pi}{3}-\cos2x \right)=\frac{1}{4}\cos2x-\frac{1}{2}\cos^{2}2x$
k) $4\cos(a-b)\cos(b-c)\cos(c-a)=2\cos(a-b)\left( \cos(b-a)+\cos(b-2c+a) \right)$
$\begin{array}{l}=2\cos(a-b)\cos(b-a)+2\cos(a-b)\cos(b-2c+a) \\ =1+\cos(2a-2b)+\cos(2a-2c)+\cos(2c-2b)\end{array}$
b) $2\cos(a+b)\cos(a-b)=\cos2a+\cos2b$
c) $4\sin3x\sin2x\cos x=2\sin3x(\sin3x+\sin x)=2\sin^{2}3x+2\sin3x\sin x=2\sin^{2}3x-\cos4x+\cos2x$
d) $4\sin\frac{13x}{2}\cos x\cos\frac{x}{2}=2\sin\frac{13x}{2}(\cos\frac{3x}{2}+\cos\frac{x}{2})=2\sin\frac{13x}{2}\cos\frac{3x}{2}+2\sin\frac{13x}{2}\cos\frac{x}{2}=$
$=\sin8x+\sin5x+\sin7x+\sin6x$.
e) $\sin(x+30^{\circ})\cos(x-30^{\circ})=\frac{1}{2}(\sin2x+\sin60^{\circ})=\frac{1}{2}\sin2x+\frac{\sqrt{3}}{4}$
f) $\sin\frac{\pi}{5}\sin\frac{2\pi}{5}=-\frac{1}{2}(\cos\frac{3\pi}{5}-\cos\frac{\pi}{5})$
g) $2\sin x\sin2x\sin3x=\sin3x(\cos x-\cos3x)=\frac{1}{2}(2\sin3x\cos x-\sin6x)=\frac{1}{2}(\sin4x+\sin2x-\sin6x)$
h) $8\cos x\sin2x\sin3x=4\sin3x(\sin3x+\sin x)=4\sin^{2}3x+4\sin3x\sin x=4\sin^{2}3x+2\cos2x-2\cos4x$
i) $\sin\left( x+\frac{\pi}{6} \right)\sin\left( x-\frac{\pi}{6} \right)\cos2x=\frac{1}{2}\cos2x\left( \cos\frac{\pi}{3}-\cos2x \right)=\frac{1}{4}\cos2x-\frac{1}{2}\cos^{2}2x$
k) $4\cos(a-b)\cos(b-c)\cos(c-a)=2\cos(a-b)\left( \cos(b-a)+\cos(b-2c+a) \right)$
$\begin{array}{l}=2\cos(a-b)\cos(b-a)+2\cos(a-b)\cos(b-2c+a) \\ =1+\cos(2a-2b)+\cos(2a-2c)+\cos(2c-2b)\end{array}$