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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 please?

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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

#1
+7336
+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

hectictar  Apr 20, 2017
#1
+7336
+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

hectictar  Apr 20, 2017
#2
+91160
+1

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

CPhill  Apr 20, 2017
edited by CPhill  Apr 20, 2017
#3
+7336
+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 !

hectictar  Apr 20, 2017
#4
+20153
+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?

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

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

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

heureka  Apr 21, 2017
edited by heureka  Apr 21, 2017
#5
+20153
+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?

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

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

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

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