Ile wynosi ten logarytm?
: 23 wrz 2017, o 14:07
Jak to ugryźć ?
\(\displaystyle{ \log_{2 \sqrt{2} } \sin \frac{3}{4} \pi}\)
\(\displaystyle{ \log_{2 \sqrt{2} } \sin \frac{3}{4} \pi}\)
\(\displaystyle{ \sin\frac{3}{4}\pi=\sin\left( \pi-\frac{\pi}{4}\right) =\sin \frac \pi 4=\frac{\sqrt{2}}{2}}\)
ponieważ \(\displaystyle{ \sin(\pi-x)=\sin x}\)
Ponadto \(\displaystyle{ \frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}=2^{-\frac 1 2}}\)
oraz \(\displaystyle{ 2\sqrt{2}=2^{\frac 3 2}}\)