Câu hỏi của nguyễn ngọc thúy vi

Question

Chứng minh các đẳng thức sau:

1/ $sin^6\alpha+cos^6\alpha=\frac{5}{8}+\frac{3}{8}cos4\alpha$

2/$\frac{1+sin2\alpha-cos2\alpha}{1+cos2\alpha}=tan\alpha+tan^2\alpha$

Nguyễn Việt Lâm · Accepted Answer

$sin^6a+cos^6a=\left(sin^2x\right)^3+\left(cos^2x\right)^3$

$=\left(sin^2x+cos^2x\right)\left(sin^4x+cos^4x-sin^2x.cos^2x\right)$

$=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2x.cos^2x$

$=\left(sin^2x+cos^2x\right)^2-\frac{3}{4}.\left(2sinx.cosx\right)^2$

$=1-\frac{3}{4}sin^22x=1-\frac{3}{4}\left(\frac{1}{2}-\frac{1}{2}cos4x\right)=\frac{5}{8}+\frac{3}{8}cos4x$

2/

$\frac{1+sin2a-cos2a}{1+cos2a}=\frac{1+2sina.cosa-\left(1-2sin^2a\right)}{1+2cos^2a-1}=\frac{2sina.cosa+2sin^2a}{2cos^2a}$

$=\frac{2sina.cosa}{2cos^2a}+\frac{2sin^2a}{2cos^2a}=tana+tan^2a$Câu trả lời: $sin^6a+cos^6a=\left(sin^2x\right)^3+\left(cos^2x\right)^3$

$=\left(sin^2x+cos^2x\right)\left(sin^4x+cos^4x-sin^2x.cos^2x\right)$

$=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2x.cos^2x$

$=\left(sin^2x+cos^2x\right)^2-\frac{3}{4}.\left(2sinx.cosx\right)^2$

$=1-\frac{3}{4}sin^22x=1-\frac{3}{4}\left(\frac{1}{2}-\frac{1}{2}cos4x\right)=\frac{5}{8}+\frac{3}{8}cos4x$

2/

$\frac{1+sin2a-cos2a}{1+cos2a}=\frac{1+2sina.cosa-\left(1-2sin^2a\right)}{1+2cos^2a-1}=\frac{2sina.cosa+2sin^2a}{2cos^2a}$

$=\frac{2sina.cosa}{2cos^2a}+\frac{2sin^2a}{2cos^2a}=tana+tan^2a$

nguyễn ngọc thúy vi · Answer

1/ x thành α nha bạnCâu trả lời: 1/ x thành α nha bạn