Question

[Quang Nguyễn]

Cho 0⁰ < x < 90⁰. Chứng minh các đẳng thức sau:

1. sin⁶ x +cos⁶ x = 1 - 3sin² x cos² x.

2. sin⁴ x - cos⁴ x = 1 - 2cos² x.

3. tan² x - sin² x = tan² x.sin²x.

4. cot² x - cos² x = cot² x.cos² x.

5.\(\left(\sqrt{\dfrac{1+sinx}{1-sinx}}-\sqrt{\dfrac{1-sinx}{1+sinx}}\right)^2\) = 4 tan² x.

6.\(\left(\sqrt{\dfrac{1+cosx}{1-cosx}}-\sqrt{\dfrac{1-cosx}{1+cosx}}\right)^2\) = 4 cot² x.

Answer 1

[Nguyễn Lê Phước Thịnh]

1: \(sin^6x+cos^6x+3sin^2x\cdot cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-3\cdot sin^2x\cdot cos^2x\cdot\left(sin^2x+cos^2x\right)+3\cdot sin^2x\cdot cos^2x\)

=1

2: \(sin^4x-cos^4x\)

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

\(=1-2\cdot cos^2x\)