Question

Sa se demonstreze ca: 
$(7+cos4x)^{2}=32(1+ sin^{8}x+ cos^{8}x)$

Multumesc! :)

Answer 1

[tex]cos2x=cos(x+x)=cos^2x-sin^2x=cos^2x-(1-sin^2x)\\ cos2x=2cos^2x-1\\ cos4x=2cos^22x-1=\\ =2(2cos^2x-1)^2-1=\\ =2(4cos^4x-4cos^2x+1)-1=\\ =8cos^4x-8cos^2x+1\\ (7+cos4x)^2=(8cos^4x-8cos^2x+8)^2=64(cos^4x-cos^2x+1)^2[1]\\ 32(1+sin^8x+cos^8x)=32[1+(sin^4x+cos^4x)^2-2sin^4xcos^4x]=\\ =32\{1+[(sin^x+cos^x)^2-2sin^xcos^x]^2-2sin^4cos^4x\}=\\ =64(1-sin^2xcos^2x)^2=64[1-(1-cos^2x)cos^2x]^2=\\ =64(1-cos^2x+cos^4x)^2[2]\\ Din [1]+[2]=>(7+cos4x)^2=32(1+sin^8x+cos^8x).[/tex]