Giải bài 3.5 trang 33

Đề bài

Chứng minh rằng:

a) ${\sin }^4\alpha +{\cos }^4\alpha =1-2{\sin }^2\alpha .{\cos }^2\alpha$.

b) ${\sin }^6\alpha +{\cos }^6\alpha =1-3{\sin }^2\alpha .{\cos }^2\alpha$.

c) $\sqrt{{\sin }^4\alpha +6{\cos }^2\alpha +3}+\sqrt{{\cos }^4\alpha +4{\sin }^2\alpha }=4$.

Lời giải chi tiết

a) ${\sin }^4\alpha +{\cos }^4\alpha =1-2{\sin }^2\alpha .{\cos }^2\alpha$.

$VT={\left({\sin }^2\alpha \right)}^2+{\left({\cos }^2\alpha \right)}^2$

$=\left({\sin }^2\alpha +{\cos }^2\alpha \right)-2{\sin }^2\alpha .{\cos }^2\alpha$

$=1-2{\sin }^2\alpha .{\cos }^2\alpha =VP$.

b) ${\sin }^6\alpha +{\cos }^6\alpha =1-3{\sin }^2\alpha .{\cos }^2\alpha$.

$VT={\left({\sin }^2\alpha \right)}^3+{\left({\cos }^2\alpha \right)}^3$

$={\left({\sin }^2\alpha +{\cos }^2\alpha \right)}^3-3{\sin }^2\alpha .{\cos }^2\alpha \left({\sin }^2\alpha +{\cos }^2\alpha \right)$

$=1-3{\sin }^2\alpha .{\cos }^2\alpha .1$

$=1-3{\sin }^2\alpha .{\cos }^2\alpha =VP$.

c) $\sqrt{{\sin }^4\alpha +6{\cos }^2\alpha +3}+\sqrt{{\cos }^4\alpha +4{\sin }^2\alpha }=4$.

$VT=\sqrt{{\left({\sin }^2\alpha \right)}^2+6{\cos }^2\alpha +3}+\sqrt{{\left({\cos }^2\alpha \right)}^2+4{\sin }^2\alpha }$

$=\sqrt{{\left(1-{\cos }^2\right)}^2+6{\cos }^2\alpha +3}+\sqrt{{\left(1-{\sin }^2\alpha \right)}^2+4{\sin }^2\alpha }$

$=\sqrt{{\cos }^4\alpha +4{\cos }^2\alpha +4}+\sqrt{{\sin }^4\alpha +2{\sin }^2\alpha +1}$

$=\sqrt{{\left({\cos }^2\alpha +2\right)}^2}+\sqrt{{\left({\sin }^2\alpha +1\right)}^2}$

$={\cos }^2\alpha +2+{\sin }^2\alpha +1=4=VP$.