求证:(1-cosa的4次方-sina的4次方)/(1-cosa的6次方-sina的6次方)=2/3
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(1-cos^4a-sin^4a)/(1-cos^6a-sin^6a)

=[1-(cos^4a+sin^4a+2cos^2asin^2a)+2cos^2asin^2a]/[1-((cos^2a)^3+(sin^2a)^3)]

=[1-(cos^2a+sin^2a)^2+2cos^2asin^2a]/[1-(cos^2a+sin^2a)(cos^4a+sin^4a-cos^2asin^2a)]

=(2cos^2asin^2a)/[1-(cos^4a+sin^4a-cos^2asin^2a)]

=2cos^2asin^2a/[1-(cos^2a+sin^2a)^2+3cos^2asin^2a]

=2cos^2asin^2a /3cos^2asin^2a

=2/3