Udowodnij, że

[Einveru]

a) jeśli c = a + b to sin^2 c = cos^ a + cos^2 b - 2 cos a cos b cos c

b) cos\(\frac{π}{5}\) cos \(\frac{2π}{5}\) = 0,25

Wygląda na coś z podwojonym kątem, ale nie wiem jak to ruszyć :/

c) cos\(\frac{π}{5}\) cos \(\frac{3π}{5}\) = - 0,25

[kerajs]

b)
\(L=cos \frac{ \pi }{5}\cos \frac{2\pi}{5}= \frac{4\sin \frac{\pi}{5} }{4\sin \frac{\pi}{5} } \cdot \cos \frac{ \pi }{5}\cos \frac{2\pi}{5}=
\frac{2 }{4\sin \frac{\pi}{5} } \cdot \sin \frac{ 2\pi }{5}\cos \frac{2\pi}{5}=\frac{1 }{4\sin \frac{\pi}{5} } \cdot \sin \frac{ 4\pi }{5}=\)
\(= \frac{1 }{4\sin \frac{\pi}{5} } \cdot \sin \left( \pi -\frac{ \pi }{5}\right)= \frac{1 }{4\sin \frac{\pi}{5} } \cdot \sin \frac{ \pi }{5}= \frac{1}{4} =P\)
c)
\(L=\cos \frac{ \pi }{5}\cos \frac{3\pi}{5}=\cos \frac{ \pi }{5}\cos \left( \pi -\frac{2}{5}\right) =-\cos \frac{ \pi }{5}\cos \frac{2\pi}{5}=\\=- \frac{4\sin \frac{\pi}{5} }{4\sin \frac{\pi}{5} } \cdot \cos \frac{ \pi }{5}\cos \frac{2\pi}{5}=-
\frac{2 }{4\sin \frac{\pi}{5} } \cdot \sin \frac{ 2\pi }{5}\cos \frac{2\pi}{5}=-\frac{1 }{4\sin \frac{\pi}{5} } \cdot \sin \frac{ 4\pi }{5}=\)
\(=- \frac{1 }{4\sin \frac{\pi}{5} } \cdot \sin \left( \pi -\frac{ \pi }{5}\right)= - \frac{1 }{4\sin \frac{\pi}{5} } \cdot \sin \frac{ \pi }{5}= \frac{-1}{4} =P\)

a)
\(\sin^2c=\sin^2(a+b)=\sin^2a\cos^2b+2\sin a\cos b\sin b\cos a+\sin^2b\cos^2a=\\=(1-\cos^2a)\cos^2b+2\sin a\cos b\sin b\cos a+(1-\cos^2b)\cos^2a=\\=\cos^2a+\cos^2b+2\sin a\cos b\sin b\cos a-2\cos^2a\cos^2b=\cos^2a+\cos^2b+2\cos a\cos b(\sin a\sin b-\cos a\cos b)=\\=\cos^2a+\cos^2b+2\cos a\cos b(-\cos (a+ b))=\cos^2a+\cos^2b-2\cos a\cos b\cos c\)

[Einveru]

b)
\(L=\cos \frac{ \pi }{5}\cos \frac{2}{5}= \frac{4\sin \frac{\pi}{5} }{4\sin \frac{\pi}{5} } \cdot \cos \frac{ \pi }{5}\cos \frac{2}{5}=
\frac{2 }{4\sin \frac{\pi}{5} } \cdot \sin \frac{ 2\pi }{5}\cos \frac{2}{5}=\frac{1 }{4\sin \frac{4\pi}{5} } \cdot \sin \frac{ 4\pi }{5}=\)
\(= \frac{1 }{4\sin \frac{4\pi}{5} } \cdot \sin \left( \pi -\frac{ \pi }{5}\right)= \frac{1 }{4\sin \frac{4\pi}{5} } \cdot \sin \frac{ \pi }{5}= \frac{1}{4} =P\)

Jak z cos \(\frac{π}{5}\) zrobiłeś sin\(\frac{2π}{5}\)? O co chodzi z sin\(\frac{4π}{5}\) i dlaczego skraczasz to z sin \(\frac{π}{5}\)?

[Galen]

\(4sin( \frac{\pi}{5}) \cdot cos( \frac{\pi}{5})=2 \cdot 2sin( \frac{\pi}{5}) cos( \frac{\pi}{5})=2 \cdot sin(2 \cdot \frac{\pi}{5})=\\=2sin( \frac{2\pi}{5})\)
Jest tu zastosowany wzór na sinus podwojonego argumentu.
Potem jeszcze raz jest w liczniku ten sam wzór...
\(2 \cdot sin( \frac{2\pi}{5})cos( \frac{2\pi}{5})=sin(2 \cdot \frac{2\pi}{5}) =sin( \frac{4\pi}{5})\)
W mianowniku też jest taki sam sinus i następuje skracanie.