Chứng minh các đẳng thức lượng giác sau: a) 4cosx.cos(pi/3-x).cos(pi/3+x)=cos3x

Giải SBT Toán 11 Bài 3: Các công thức lượng giác

Bài 4 trang 19 SBT Toán 11 Tập 1: 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) $\frac{\sin 2x\cos x}{\left(1+\cos x\right)\left(1+\cos 2x\right)}=\tan \frac{x}{2};$

c) sinx(1 + 2cos2x + 2cos4x + 2cos6x) = sin7x;

d) $\frac{{\sin }^{2}3x}{{\sin }^{2}x}-\frac{{\cos }^{2}3x}{{\text{cos}}^{2}x}=8\cos 2x.$

Lời giải:

a) $4\text{cos}x\text{cos}\left(\frac{\pi }{3}-x\right)\text{cos}\left(\frac{\pi }{3}+x\right)=2\text{cos}x\left(\text{cos}2x+\text{cos}\frac{2\pi }{3}\right)$

$=2\text{cos}x\text{cos}2x+2\text{cos}x\text{cos}\frac{2\pi }{3}$

$=\text{cos}x+\text{cos}3x+2\text{cos}x\cdot \left(-\frac{1}{2}\right)$

$=\text{cos}x+\text{cos}3x-\text{cos}x=\text{cos}3x.$

b) $\frac{\text{sin}2x\text{cos}x}{\left(1+\text{cos}x\right)\left(1+\text{cos}2x\right)}=\frac{\left(2\text{sin}x\text{cos}x\right)\text{cos}x}{\left(1+2{\text{cos}}^2\frac{x}{2}-1\right)\left(1+2{\text{cos}}^{2}x-1\right)}$

$=\frac{2\text{sin}x{\text{cos}}^{2}x}{4{\text{cos}}^2\frac{x}{2}{\text{cos}}^{2}x}$

$=\frac{\text{sin}x}{2{\text{cos}}^2\frac{x}{2}}=\frac{2\text{sin}\frac{x}{2}\text{cos}\frac{x}{2}}{2{\text{cos}}^2\frac{x}{2}}=\frac{\text{sin}\frac{x}{2}}{\text{cos}\frac{x}{2}}=\text{tan}\frac{x}{2}.$

c) sinx(1 + 2cos2x + 2cos4x + 2cos6x)

= sinx + 2sinxcos2x + 2sinxcos4x + 2sinxcos6x

= sinx + [sin(‒x) + sin3x] + [sin(‒3x) + sin5x] + [sin(‒5x) + sin7x]

= sinx + (‒sinx + sin3x) + (‒sin3x + sin5x) + (‒sin5x + sin7x)

= sin7x.

d)

$\frac{{\text{sin}}^{2}3x}{{\text{sin}}^{2}x}-\frac{{\text{cos}}^{2}3x}{{\text{cos}}^{2}x}=\frac{{\text{sin}}^{2}3x{\text{cos}}^{2}x-{\text{cos}}^{2}3x{\text{sin}}^{2}x}{{\text{sin}}^{2}x{\text{cos}}^{2}x}=\frac{{\left(\text{sin}3x\text{cos}x\right)}^2-{\left(\text{cos}3x\text{sin}x\right)}^2}{{\text{sin}}^{2}x{\text{cos}}^{2}x}$

$=\frac{\left(\text{sin}3x\text{cos}x+\text{cos}3x\text{sin}x\right)\left(\text{sin}3x\text{cos}x-\text{cos}3x\text{sin}x\right)}{\frac{1}{4}{\text{sin}}^{2}2x}$

$=\frac{4\text{sin}4x\text{sin}2x}{{\text{sin}}^{2}2x}=\frac{4\left(2\text{sin}2x\text{cos}2x\right)\text{sin}2x}{{\text{sin}}^{2}2x}$

$=\frac{8{\text{sin}}^{2}2x\text{cos}2x}{{\text{sin}}^{2}2x}=8\text{cos}2x.$