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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?

Guest Apr 20, 2017

Best Answer 

 #1
avatar+5541 
+3

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

 

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 smiley

hectictar  Apr 20, 2017
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5+0 Answers

 #1
avatar+5541 
+3
Best Answer

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

 

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 smiley

hectictar  Apr 20, 2017
 #2
avatar+79645 
+1

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

 

 

cool cool cool

CPhill  Apr 20, 2017
edited by CPhill  Apr 20, 2017
 #3
avatar+5541 
+3

AH, I know you would have seen that before long..

What's really ingenious is how you figured out that MrSilvers was trying to take the sin of 15ยบ ...That took some real insight ! smiley

hectictar  Apr 20, 2017
 #4
avatar+18764 
+2

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?

 

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

 

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

 

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

 

 

laugh

heureka  Apr 21, 2017
edited by heureka  Apr 21, 2017
 #5
avatar+18764 
+1

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?

 

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

 

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

 

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

 

 

laugh

heureka  Apr 21, 2017
edited by heureka  Apr 21, 2017

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