Câu hỏi 5 - Mục Bài tập trang 14

1. Nội dung câu hỏi

Chứng minh các đẳng thức lượng giác sau:

a) \({\sin ^4}x + {\cos ^4}x = 1 - 2{\sin ^2}x{\cos ^2}x\);

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

3. Lời giải chi tiết 

a) \({\sin ^4}x + {\cos ^4}x = {\sin ^4}x + 2{\sin ^2}x{\cos ^2}x + {\cos ^4}x - 2{\sin ^2}x{\cos ^2}x\)

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

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

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

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

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

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