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Question

Chứng minh các biểu thức sau không phụ thuộc vào

$x$ :

$A=\sin^4x(3-2\sin^2x)+\cos^4x(3-2\cos^2x)$

$B=3(\sin^8x-\cos^8x)+4(\cos^6x-2\sin^6x)+6\sin^4x$

$C=\cos^2x+\cos^2(a+x)-2\cos a\cos x\cos(a+x)$

$D=\sin^2(a+x)+\sin^2(a-x)+2\sin(a+x)\sin(a-x)\cos2a$

score 4 · Answer 1

$D=\sin^2(a-x)+\sin(a+x)[\sin(a+x)+2\sin(a-x)\cos2a]$

$=\sin^2(a-x)+\sin(a+x)[\sin(a+x)+\sin(3a-x)+\sin(-a-x)]$

$=\sin^2(a-x)+\sin(a+x)\sin(3a-x)$

$=\frac{1-\cos(2a-2x)}{2}+\frac{1}{2}\cos(-2a+2x)-\frac{1}{2}\cos4a$

$=\frac{1}{2}(1-\cos4a)=\sin^22a$

score 4 · Answer 2

$C=\cos^2x+\cos(a+x)[\cos(a+x)-2\cos a\cos x]$

$=\cos^2x+\cos(a+x)[-\cos a\cos x-\sin a\sin x]$

$=\cos^2x-\cos(a+x)\cos(a-x)$

$=\frac{1}{2}+\frac{1}{2}\cos2x-\frac{1}{2}(\cos2a+\cos2x)$

$=\sin^2a$

score 4 · Answer 3

$B=3[a^4-(1-a)^4]+4[(1-a)^3-2a^3]+6a^2$ , với

$a=\sin^2x$

$=3(4a^3-6a^2+4a-1)+4(-3a^3+3a^2-3a+1)+6a^2$

$=12a^3-18a^2+12a-3-12a^3+12a^2-12a+4+6a^2=1$

score 4 · Answer 4

$A=3(\sin^4x+\cos^4x)-2(\sin^6x+\cos^6x)$

$=3[(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x]-2(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)$

$=3(1-\frac{1}{2}\sin^22x)-2(1-\frac{3}{4}\sin^22x)=1$

4 Đáp án