Question

Exercitiul 2, repede va rog. Multumesc

Answer 1

$a)\:\sin^2{x}+\sin^2{x}\cdot ctg^2{x}=\sin^2{x}+\sin^2{x}\cdot\frac{\cos^2{x}}{\sin^2{x}}=\sin^2{x}+\cos^2{x}=1$

$b)\:ctg^2{x}\cdot(tg^2{x}-\sin^2{x})=\frac{\cos^2{x}}{\sin^2{x}}\cdot(\frac{\sin^2{x}}{\cos^2{x}}-\frac{\sin^2x\cos^2x}{\cos^2x})=\\=\frac{\cos^2{x}}{\sin^2{x}}\cdot\frac{\sin^2x(1-\cos^2x)}{\cos^2x}=1-\cos^2x=\sin^2x$

$c)\:a^2-b^2=(a+b)(a-b)\\\cos^4x-\sin^4x=(\cos^2x+\sin^2x)(\cos^2x-\sin^2x)=\cos^2x-\sin^2x=\\=\cos^2x-(1-\cos^2x)=2\cos^2x-1$

$d)\:(a+b)^2=a^2+2ab+b^2\leftrightarrow a^2+b^2=(a+b)^2-2ab\\\cos^4x+\sin^4x=(\cos^2x+\sin^2x)^2-2\cos^2x\sin^2x=1-2\cos^2x\sin^2x$

$e)\:(a+b)^3=a^3+b^3+3a^2b+3ab^2\leftrightarrow a^3+b^3=(a+b)^3-\\-3a^2b-3ab^2\\\cos^6x+\sin^6x=(\cos^2x+\sin^2x)^3-3\cos^4x\sin^2x-3\cos^2x\sin^4x=\\=1-3\cos^2x\sin^2x(\cos^2x+\sin^2x)=1-3\cos^2x\sin^2x$

$f)\:(a-b)^3=a^3-b^3-3a^2b+3ab^2\leftrightarrow a^3-b^3=(a-b)^3+3a^2b-3ab^2\\\cos^6x-\sin^6x=(\cos^2x-\sin^2x)^3+3\cos^4x\sin^2x-3\cos^2x\sin^4x=\\=(\cos^2x-\sin^2x)^3+3\cos^2x\sin^2x(\cos^2x-\sin^2x)=\\=(\cos^2x-\sin^2x)((\cos^2x-\sin^2x)^2+3\cos^2x\sin^2x)=\\=(1-\sin^2x-\sin^2x)(\cos^4x-2\sin^2x\cos^2x+\sin^4x+3\cos^2x\sin^2x)\\\text{folosim punctul d)}\\=(1-2\sin^2x)(1-2\cos^2x\sin^2x-2\cos^2x\sin^2x+3\cos^2x\sin^2x)=\\=(1-2\sin^2x)(1-\cos^2x\sin^2x)$