Question

Câu 1 : Chứng minh rằng : 3 - 4sin²x = 4cos²x - 1Câu 2 : Chứng minh rằng : cos⁴x - sin⁴x = 2cos²x - 1 = 1 - 2sin²xCâu 3 : Chứng minh rằng : sin⁴x + cos⁴x = 1 - 2sin²xCos²x

Answer 1

1/ \(3-4\sin^2=4\cos^2x-1\Leftrightarrow4\left(\sin^2x+\cos^2x\right)-4=0\Leftrightarrow4.1-4=0\left(ld\right)\Rightarrow dpcm\)

2/ \(\cos^4x-\sin^4x=\left(\cos^2x+\sin^2x\right)\left(\cos^2x-\sin^2x\right)=\cos^2x-\left(1-\cos^2x\right)=2\cos^2x-1=\left(1-\sin^2x\right)-\sin^2x=1-2\sin^2x\)

3/ \(\sin^4x+\cos^4x=\left(\sin^2x+\cos^2x\right)^2-2\sin^2x.\cos^2x=1-2\sin^2x.\cos^2x\)