Chứng minh hệ thức trong tam giác không phụ thuộc vào $\rm \alpha$. $\rm \ a) \ A=3(sin^4\alpha+cos^4\alpha)-2(sin^6\alpha+cos^6\alpha)$ $\rm \ b) \ B

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1. a,

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

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

   $=-2\sin^4\alpha+2\sin^2\alpha\cos^2\alpha-2\cos^4\alpha+3\sin^4\alpha+3\cos^4\alpha$

   $=(\sin^2\alpha)^2+2\sin^2\alpha\cos^2\alpha+(\cos^2\alpha)^2$

   $=(\sin^2\alpha+\cos^2\alpha)^2$

   $=1^2$

   $=1$

   b,

   $B=\sin^6a+\cos^6\alpha+3\sin^2\alpha\cos^2\alpha$

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

   $=(\sin^2\alpha)^2+2\sin^2\alpha\cos^2\alpha+(\cos^2\alpha)^2$

   $=(\sin^2\alpha+\cos^2\alpha)^2$

   $=1^2$

   $=1$

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