Question

sa se calculeze cos^8 (5pi/12) - cos^8(pi/12)

Answer 1

   
[tex]\displaystyle\\ \cos^8 \Big(\frac{5\pi}{12} \Big) - \cos^8 \Big(\frac{\pi}{12} \Big)\\\\ \text{Transformam radianii in grade.}\\\\ \frac{\pi}{12} = \frac{180}{12}=15^o\\\\ \frac{5\pi}{12} = \frac{5\times 180}{12} = 5\times 15 = 75^o\\\\ \Longrightarrow~~\cos^8 75^o- \cos^8 15^o\\\\ 75^o+15^o = 90^o\\\\ \Longrightarrow~~\cos 75^o = sin 15^o\\\\ \Longrightarrow~~\cos^8 75^o- \cos^8 15^o = \sin^8 15^o- \cos^8 15^o[/tex]


[tex]\displaystyle\\ \text{Aplicam formulele:}\\\\ \sin \frac{\alpha}{2} = \pm \sqrt{ \frac{1-\cos \alpha}{2}} ~~~\text{ si }~~~ \cos \frac{\alpha}{2} = \pm \sqrt{ \frac{1+\cos \alpha}{2}} \\\\ \texttt{Rezulta: }\\\\ \sin15^o = \sin \frac{30^o}{2} = \pm \sqrt{ \frac{1-\cos 30^o}{2}}\\\\ \cos 15^o=\cos\frac{30^o}{2} = \pm \sqrt{ \frac{1+\cos 30^o}{2}} \\\\ \Longrightarrow~\sin^8 15^o- \cos^8 15^o = \left(\pm \sqrt{ \frac{1-\cos 30^o}{2}}\right)^8 - \left(\pm \sqrt{ \frac{1+\cos 30^o}{2}}\right)^8= [/tex]


[tex]\displaystyle\\ = \left(\frac{1-\cos 30^o}{2}\right)^4 - \left( \frac{1+\cos 30^o}{2}\right)^4=\\\\ =\left(\frac{1- \dfrac{ \sqrt{3} }{2} }{2}\right)^4 - \left(\frac{1+ \dfrac{ \sqrt{3} }{2} }{2}\right)^4=\left(\frac{ \dfrac{2- \sqrt{3} }{2} }{2}\right)^4 - \left(\frac{\dfrac{ 2+\sqrt{3} }{2} }{2}\right)^4=\\\\\\ = \left(\dfrac{2- \sqrt{3} }{4}\right)^4-\left(\dfrac{2+ \sqrt{3} }{4}\right)^4= \left(\dfrac{(2- \sqrt{3})^2 }{16}\right)^2-\left(\dfrac{(2+ \sqrt{3})^2 }{16}\right)^2=\\\\ [/tex]


[tex]\displaystyle\\ =\left(\dfrac{7- 4\sqrt{3} }{16}\right)^2-\left(\dfrac{7+ 4\sqrt{3} }{16}\right)^2=\\\\ =\dfrac{49- 56\sqrt{3}+48 }{256}-\dfrac{49+ 56\sqrt{3}+48 }{256}=\\\\ =\dfrac{97- 56\sqrt{3} }{256}-\dfrac{97+ 56\sqrt{3} }{256}=\\\\ =\dfrac{97- 56\sqrt{3} -(97+ 56\sqrt{3}) }{256}=\\\\ =\dfrac{97- 56\sqrt{3} -97- 56\sqrt{3} }{256}=\\\\ =\dfrac{- 112\sqrt{3}}{256}=-\dfrac{^{16)}112\sqrt{3}}{256}= \boxed{\bf\frac{7 \sqrt{3}}{16}} [/tex]