СРОЧНО. ДАЮ МНОГО БАЛЛОВ. Created by szplas. algebra-ru

СРОЧНО. ДАЮ МНОГО БАЛЛОВ...

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√2 sin(п/16) cos(п/16) ( cos²(П/16) - sin² (П/16) )= √2/2 sin (П/8) cos (П/8)=
√2/4 Sin(П/4)=√2/4 * √2/2=1/4

log√2(cos(П/8) ) +log√2(cos(3П/8) ) =log√2(cos(П/8)сos(3П/8) )=
log√2 (1/2 (cos (3П /8+П/8) + Cos (3П/8-п/8) ) )=
log√2 (1/2 (cos (П/2) + Cos (П/4) ) )=log√2 (1/2 cos (П/4) )=log√2 (1/2 *√2/2 )=
log√2 (√2/4)=log√2 (√2/(√2*√2*√2*√2)=og√2 (1/(√2)³)=og√2 (√2^(-3))=-3

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

$\mathtt{\log_{\sqrt{2}}cos\frac{\pi}{8}+\log_{\sqrt{2}}cos\frac{3\pi}{8}=\log_{\sqrt{2}}cos\frac{\pi}{8}cos\frac{3\pi}{8}=}\\\mathtt{\log_{\sqrt{2}}\frac{cos(\frac{\pi}{8}-\frac{3\pi}{8})+cos(\frac{\pi}{8}+\frac{3\pi}{8})}{2}=\log_{\sqrt{2}}(cos\frac{\pi}{4}+cos\frac{\pi}{2})-\log_{\sqrt{2}}2=}\\\mathtt{\log_{\sqrt{2}}(\frac{\sqrt{2}}{2}+0)-2=\log_{\sqrt{2}}\frac{\sqrt{2}}{2}-2=\log_{\sqrt{2}}\sqrt{2}-\log_{\sqrt{2}}2-2=-3}$

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