Chứng minh các đẳng thức lượng giác sau: a) sin4x + cos4x = 1 ‒ 2sin2xcos2x

Giải SBT Toán 11 Bài 2: Giá trị lượng giác của một góc lượng giác

Bài 5 trang 14 SBT Toán 11 Tập 1: Chứng minh các đẳng thức lượng giác sau:

a) sin⁴x + cos⁴x = 1 ‒ 2sin²xcos²x.

b) $\frac{1+\cot x}{1-\cot x}=\frac{\tan x+1}{\tan x-1};$

c) $\frac{\sin \alpha +\cos \alpha }{{\sin }^3\alpha }=\frac{1-{\cot }^4\alpha }{1-\cot \alpha };$

d) $\frac{{\tan }^2\alpha +{\cos }^2\alpha -1}{{\cot }^2\alpha +{\sin }^2\alpha -1}={\tan }^6\alpha .$

Lời giải:

a)

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

b) $\frac{1+\text{cot}x}{1-\text{cot}x}=\frac{1+\frac{1}{\text{tan}x}}{1-\frac{1}{\text{tan}x}}=\frac{\frac{\text{tan}x+1}{\text{tan}x}}{\frac{\text{tan}x-1}{\text{tan}x}}=\frac{\text{tan}x+1}{\text{tan}x-1}.$

c)

$\frac{\text{sin}\alpha +\text{cos}\alpha }{{\text{sin}}^3\alpha }=\frac{1}{{\text{sin}}^2\alpha }+\frac{\text{cos}\alpha }{\text{sin}\alpha }\cdot \frac{1}{{\text{sin}}^2\alpha }=\left(1+{\text{cot}}^2\alpha \right)+\text{cot}\alpha \left(1+{\text{cot}}^2\alpha \right)$

$=\left(1+\text{cot}\alpha \right)\left(1+{\text{cot}}^2\alpha \right)$

$=\frac{\left(1-{\text{cot}}^2\alpha \right)\left(1+{\text{cot}}^2\alpha \right)}{1-\text{cot}\alpha }=\frac{1-{\text{cot}}^4\alpha }{1-\text{cot}\alpha }$

d) $\frac{{\text{tan}}^2\alpha +{\text{cos}}^2\alpha -1}{{\text{cot}}^2\alpha +{\text{sin}}^2\alpha -1}=\frac{{\text{tan}}^2\alpha -{\text{sin}}^2\alpha }{{\text{cot}}^2\alpha -{\text{cos}}^2\alpha }=\frac{\frac{{\text{sin}}^2\alpha }{{\text{cos}}^2\alpha }-{\text{sin}}^2\alpha }{\frac{{\text{cos}}^2\alpha }{{\text{sin}}^2\alpha }-{\text{cos}}^2\alpha }$

$=\frac{{\text{sin}}^2\alpha \left(\frac{1}{{\text{cos}}^2\alpha }-1\right)}{{\text{cos}}^2\alpha \left(\frac{1}{{\text{sin}}^2\alpha }-1\right)}={\text{tan}}^2\alpha \cdot \frac{{\text{tan}}^2\alpha }{{\text{cot}}^2\alpha }={\text{tan}}^6\alpha$