Question

Помогите пожалуйста с решением, никак понять не могу

Answer 1

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




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