Chứng minh các đẳng thức lượng giác sau: a) sin^2 (x+pi/8)-sin^2(x-pi/8)=căn 2/2 .sin 2x

Giải SBT Toán 11 Bài tập cuối chương 1 trang 32

Bài 3 trang 34 SBT Toán 11 Tập 1: Chứng minh các đẳng thức lượng giác sau:

a) ${\sin }^2\left(x+\frac{\pi }{8}\right)-{\sin }^2\left(x-\frac{\pi }{8}\right)=\frac{\sqrt{2}}{2}\sin 2x;$

b) sin²y + 2cosxcosycos(x ‒ y) = cos²x + cos²(x ‒ y).

Lời giải:

a) ${\text{sin}}^2\left(x+\frac{\pi }{8}\right)-{\text{sin}}^2\left(x-\frac{\pi }{8}\right)$

$=\left.\text{sin}\left(x+\frac{\pi }{8}\right)+\text{sin}\left(x-\frac{\pi }{8}\right)\right]\left.\text{sin}\left(x+\frac{\pi }{8}\right)-\text{sin}\left(x-\frac{\pi }{8}\right)\right]$

$=\left(2\text{sin}x\text{cos}\frac{\pi }{8}\right)\left(2\text{cos}x\text{sin}\frac{\pi }{8}\right)=\left(2\text{sin}x\text{cos}x\right)\left(2\text{cos}\frac{\pi }{8}\text{sin}\frac{\pi }{8}\right)$

$=\text{sin}2x\text{sin}\frac{\pi }{4}=\frac{\sqrt{2}}{2}\text{sin}2x$

b) sin²y + 2cosxcosycos(x ‒ y) = cos²x + cos²(x ‒ y).

⇔ 2cosxcosycos(x ‒ y) ‒ cos²(x ‒ y) = cos²x ‒ sin²y

Ta có:

VT = 2cosxcosycos(x ‒ y) ‒ cos²(x ‒ y)

= cos(x – y)[2cosxcosy – cos(x – y)]

= cos(x – y)[2cosxcosy – (cosxcosy + sinxsiny)]

= cos(x – y)(cosxcosy – sinxsiny)

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

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

Vậy sin²y + 2cosxcosycos(x ‒ y) = cos²x + cos²(x ‒ y).