cos^2(pi/8)+cos^2(3pi/8)+cos^2(5pi/8)+cos^2(7pi/8) can you solve it showing me the whole way of how doing it please?

\(\cos^2(\frac{\pi}{8})+\cos^2(\frac{3\pi}{8})+\cos^2(\frac{5\pi}{8})+\cos^2(\frac{7\pi}{8}) \\~\\ \cos^2(\frac12\cdot\frac{\pi}{4})+\cos^2(\frac12\cdot\frac{3\pi}{4})+\cos^2(\frac12\cdot\frac{5\pi}{4})+\cos^2(\frac12\cdot\frac{7\pi}{4})\)

Apply the half-angle formula: \( \cos^2(\frac12\cdot a)=\frac12(1+\cos a) \)

\(\frac12(1+\cos\frac{\pi}{4})+\frac12(1+\cos \frac{3\pi}{4})+\frac12(1+\cos\frac{5\pi}{4})+\frac12(1+\cos\frac{7\pi}{4})\)

Now we can evaluate the cosines.

\(\frac12(1+\frac{\sqrt2}{2})+\frac12(1-\frac{\sqrt2}{2})+\frac12(1-\frac{\sqrt2}{2})+\frac12(1+\frac{\sqrt2}{2})\)

Simplify.

\((1+\frac{\sqrt2}{2})+(1-\frac{\sqrt2}{2}) \\~\\ 1+\frac{\sqrt2}{2}+1-\frac{\sqrt2}{2} \\~\\ 2\)

If you want more steps shown please just say so 

Very ingenious, hectictar.....it would not have occurred to me  to apply the half-angle formula!!!!

cos^2(pi/8)+cos^2(3pi/8)+cos^2(5pi/8)+cos^2(7pi/8)

can you solve it showing me the whole way of how doing it?

\(\begin{array}{l} \cos^2(\frac18\pi)+\cos^2(\frac38\pi)+\cos^2(\frac58\pi)+\cos^2(\frac78\pi) =\ ? \\ \end{array}\)

\(\begin{array}{|rcll|} \hline \cos(\frac78\pi) = \cos(\pi - \frac18\pi) = -\cos(\frac18\pi) \\ \cos(\frac58\pi) = \cos(\pi - \frac38\pi) = -\cos(\frac38\pi) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \cos^2(\frac18\pi)+\cos^2(\frac38\pi)+\cos^2(\frac58\pi)+\cos^2(\frac78\pi) \\ &=& \cos^2(\frac18\pi)+\cos^2(\frac38\pi) + \Big[ -\cos(\frac38\pi) \Big]^2 + \Big[ -\cos(\frac18\pi) \Big]^2 \\ &=& \cos^2(\frac18\pi)+\cos^2(\frac38\pi) + \cos^2(\frac38\pi) + \cos^2(\frac18\pi) \\ &=& 2\cdot \Big( \cos^2(\frac18\pi)+\cos^2(\frac38\pi) \Big) \quad | \quad \cos(\frac38\pi) = \sin(\frac12\pi-\frac38\pi) = \sin(\frac18\pi) \\ &=& 2\cdot \Big( \underbrace{ \cos^2(\frac18\pi)+\sin^2(\frac18\pi) }_{=1} \Big) \\ &=& 2 \\ \hline \end{array}\)

cos^2(pi/8)+cos^2(3pi/8)+cos^2(5pi/8)+cos^2(7pi/8)
can you solve it showing me the whole way of how doing it?

\(\begin{array}{l} \cos^2(\frac18\pi)+\cos^2(\frac38\pi)+\cos^2(\frac58\pi)+\cos^2(\frac78\pi) = ? \\ \end{array}\)

\(\begin{array}{|lcll|} \hline \cos(\frac38\pi) = \sin(\frac12\pi-\frac38\pi) = \sin(\frac18\pi) \\ \cos(\frac58\pi) = \sin(\frac12\pi-\frac58\pi) = -\sin(\frac18\pi) = -\sin(\pi -\frac18\pi ) = -\sin(\frac78 \pi) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \cos^2(\frac18\pi)+\cos^2(\frac38\pi)+\cos^2(\frac58\pi)+\cos^2(\frac78\pi) \\ &=& \cos^2(\frac18\pi)+ \Big[ \sin(\frac18\pi) \Big]^2 + \Big[ -\sin(\frac78 \pi) \Big]^2 +\cos^2(\frac78\pi) \\ &=& \underbrace{ \cos^2(\frac18\pi)+ \sin^2(\frac18\pi) }_{=1} + \underbrace{ \sin^2(\frac78 \pi) +\cos^2(\frac78\pi) }_{=1} \\ &=& 1+1 \\ &=& 2 \\ \hline \end{array} \)