Question

упростить выражение

$sin^6alpha + cos^6 alpha + frac{3-3cos^2alpha}{1+tg^2alpha}$

Answer 1

$1+tan^2 alpha=1/cos^2 alpha$

$3-3cos^2 alpha=3(1-cos^2 alpha)=3sin^2alpha$

 

$sin^6alpha+cos^6alpha+dfrac{3-3cos^2alpha}{1+tan^2alpha}=sin^6alpha+(1-sin^2alpha)^3+3sin^2alphacos^2alpha=$

$=sin^6alpha+1-3sin^2alpha+3sin^4alpha-sin^6alpha+3sin^2alphacos^2alpha=$

$=1-3sin^2alpha(1-sin^2alpha)+3sin^2alphacos^2alpha=1-3sin^2alphacos^2alpha+3sin^2alphacos^2alpha=1$