Question

Помогите пожалуйста с решением, никак понять не могу

Answer 1

$3(log_8(1+cosfrac{pi}{4})-log_{frac{1}{8}}(1-cosfrac{pi}{4}))=\=3(log_8(1+cosfrac{pi}{4})+log_{8}(1-cosfrac{pi}{4}))=\=3(log_8((1+cosfrac{pi}{4})(1-cosfrac{pi}{4})))=3(log_8(1-cos^2frac{pi}{4}))=\=3log_8(sin^2frac{pi}{4})=3log_{2^3}(frac{sqrt{2}}{2})^2=3*frac{1}{3}*log_2frac{1}{2}=log_2(2^{-1})=-1$




$log_{16}(1+cosfrac{pi}{8})+log_{16}(1-cosfrac{pi}{8})+2log_{16}(cosfrac{pi}{8})=\=log_{16}((1+cosfrac{pi}{8})(1-cosfrac{pi}{8}))+log_{16}(cos^2frac{pi}{8})=\=log_{16}(1-cos^2frac{pi}{8})+log_{16}(cos^2frac{pi}{8})=log_{16}(sin^2frac{pi}{8})+log_{16}(cos^2frac{pi}{8})=\=log_{16}(sin^2frac{pi}{8}*cos^2frac{pi}{8})=log_{16}(frac{1}{2}*2sinfrac{pi}{8}cosfrac{pi}{8})^2=log_{16}(frac{1}{4}*sin^2frac{pi}{4})=$
$=log_{16}(frac{1}{4}*(frac{sqrt{2}}{2})^2)=log_{16}frac{1}{8}=log_{2^4}2^{-3}=-frac{3}{4}$

Answer 2

спасибо большое