+0  
 
0
138
6
avatar

cos(pi/7) + (cos(pi/7))^2 - 2(cos(pi/7))^3

Guest Jul 15, 2017

Best Answer 

 #4
avatar+18612 
+2

cos(pi/7) + (cos(pi/7))^2 - 2(cos(pi/7))^3 = ?

step by step

 

\(\begin{array}{|l|rcll|} \hline \mathbf{1. \text{ Substitution } } \\ x = \frac{\pi}{7} && \cos(\frac{\pi}{7}) + \cos^2(\frac{\pi}{7}) - 2\cos^3(\frac{\pi}{7}) \\ &=& \cos(x) + \cos^2(x) - 2\cos^3(x) \\ \hline \end{array}\)

 

\(\mathbf{2. \cos^2(x) \text{ to } \cos(2x) \text{ and } \cos^3(x) \text{ to } \cos(3x) }\)

\(\begin{array}{|rcll|} \hline \text{Formula}: \\ \cos(2x) &=& 2\cos^2(x) - 1 \qquad \text{ or } \qquad \cos^2(x) = \frac12\Big(1+\cos(2x)\Big) \\ \cos(3x) &=& 4\cos^3(x) - 3\cos(x) \qquad \text{ or } \qquad \cos^3(x) = \frac14\Big(3\cos(x)+\cos(3x)\Big) \\ \\ && \cos(x) + \cos^2(x) - 2\cos^3(x) \\ &=& \cos(x) + \frac12\Big(1+\cos(2x)\Big) - \frac24\Big(3\cos(x)+\cos(3x)\Big) \\ &=& \frac12\Big( 2\cos(x) + 1 + \cos(2x)-3\cos(x) - \cos(3x) \Big) \\ &=& \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ \hline \end{array} \)

 

3. Change minus to plus

\(\begin{array}{|l|rclcl|} \hline \text{Formula}: \\ \cos(x) = -\cos(\pi-x) \\\\ & \cos(\frac{\pi}{7}) &=& -\cos(\pi-\frac{\pi}{7}) \\ & &=& -\cos(\frac67 \pi) \\ & \mathbf{\cos(x)} &\mathbf{=}&\mathbf{ -\cos(6x)} \\\\ & \cos(\frac{3\pi}{7}) &=& -\cos(\pi-\frac{3\pi}{7}) \\ & &=& -\cos(\frac47 \pi) \\ & \mathbf{\cos(3x) } &\mathbf{=}& \mathbf{-\cos(4x)} \\\\ \hline \end{array} \\ \begin{array}{|rcll|} \hline && \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ &=& \frac12\Big( 1+ \cos(6x)+\cos(2x)+\cos(4x) \Big) \\ &=& \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ \hline \end{array} \)

 

\(\begin{array}{|l|rcll|} \hline \mathbf{4. \text{ Substitution } } \\ y = 2x && \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ &=& \frac12\Big( 1+\cos(y)+\cos(2y)+ \cos(3y) \Big) \\ \hline \end{array} \)

 

5. complex numbers \(1+\cos(y)+\cos(2y)+ \cos(3y) = \ ?\)

\(\begin{array}{|rcll|} \hline C &=& 1+\cos(y)+\cos(2y)+ \cos(3y) \\ S &=& \qquad \sin(y)+\sin(2y)+ \sin(3y) \\ \mathbf{T} &\mathbf{=}& \mathbf{ C+i\cdot S } \\\\ T &=& 1+\cos(y)+\cos(2y)+ \cos(3y)+i\cdot \Big( \sin(y)+\sin(2y)+ \sin(3y) \Big)\\ &=& 1+\underbrace{\cos(y)+i\cdot \sin(y)}_{=e^{iy}} +\underbrace{\cos(2y)+i\cdot \sin(2y)}_{=e^{i2y}} +\underbrace{ \cos(3y)+i\cdot \sin(3y)}_{=e^{i3y}} \\ &=& 1+ e^{iy} + e^{i2y} + e^{i3y} \\ &=& 1+ e^{iy} + (e^{iy})^2 + (e^{iy})^3 \qquad \text{geometric series } r=e^{iy} \\ T &=& 1+ r + r^2 + r^3 \\ \hline T &=& 1+ r + r^2 + r^3 \\ r\cdot T &=& \qquad r + r^2 + r^3 + r^4 \\ \hline r\cdot T - T &=& r^4 - 1 \\ T\cdot (r-1) &=& r^4 - 1 \\ T &=& \frac{r^4 - 1} {r-1} \\ &=& \frac{(e^{iy})^4 - 1} {e^{iy}-1} \\ \mathbf{T} & \mathbf{=} & \mathbf{\frac{e^{i4y} - 1} {e^{iy}-1}} \\ \hline \end{array}\)

 

\(\begin{array}{|l|rclcl|} \hline \text{Formula}: \\ e^{i\phi} - 1 = 2i\cdot \sin(\frac{\phi}{2})\cdot e^{i\frac{\phi}{2}} \\\\ & \mathbf{T} & \mathbf{=} & \mathbf{\frac{e^{i4y} - 1} {e^{iy}-1}} \\ & &=& \frac{ 2i\cdot \sin(2y) \cdot e^{i2y} } { 2i\cdot \sin(\frac{y}{2})\cdot e^{ i\frac{y}{2} }} \qquad | \qquad y = 2x \\ & &=& \frac{ \sin(4x) \cdot e^{i4x} } { \sin(x)\cdot e^{ ix }} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot e^{i4x-ix} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot e^{i3x} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot \Big( \cos(3x) + i\cdot \sin(3x) \Big) \\ & &=& \underbrace{\frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x)}_{=C} + i\cdot \underbrace{ \frac{ \sin(4x) } { \sin(x)} \cdot \sin(3x) }_{=S} \\\\ \hline \end{array}\)

 

6. Solution

\(\begin{array}{|rcll|} \hline & C &=& 1+\cos(y)+\cos(2y)+ \cos(3y) \qquad | \qquad y = 2x \\ & &=& 1+\cos(2x)+\cos(4x)+ \cos(6x) \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x) \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \text{Formula}: \\ \sin(x) = \sin(\pi-x) \\\\ & \cos(\frac{4\pi}{7}) &=& \sin(\pi-\frac{4\pi}{7}) \\ & &=& \sin(\frac37 \pi) \\ & \mathbf{\sin(4x)} &\mathbf{=}&\mathbf{ \sin(3x)} \\\\ & \cos(\frac{6\pi}{7}) &=& \sin(\pi-\frac{6\pi}{7}) \\ & &=& \sin(\frac17 \pi) \\ & \mathbf{\sin(6x)} &\mathbf{=}&\mathbf{ \sin(x)} \\\\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \cos(\frac{\pi}{7}) + \cos^2(\frac{\pi}{7}) - 2\cos^3(\frac{\pi}{7}) \quad & | \quad x=\frac{\pi}{7} \\ &=& \cos(x) + \cos^2(x) - 2\cos^3(x) \\ &=& \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ &=& \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ &=& \frac12\Big( \frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x) \Big) \quad & | \quad \mathbf{\sin(4x)= \sin(3x)}\\ &=& \frac12\Big( \frac{ \sin(3x) } { \sin(x)} \cdot \cos(3x) \Big) \quad & | \quad \sin(3x)\cos(3x)=\frac{\sin(6x)}{2} \\ &=& \frac12\Big( \frac{ \sin(6x) } { 2 \sin(x)} \Big) \quad & | \quad \mathbf{\sin(6x)=\sin(x)} \\ &=& \frac12\Big( \frac{ \sin(x) } { 2 \sin(x)} \Big) \\ &=& \frac12\Big( \frac12 \Big) \\ &=& \mathbf{\frac14} \\ \hline \end{array}\)

 

 

laugh

heureka  Jul 18, 2017
Sort: 

6+0 Answers

 #1
avatar+90567 
+1

Maybe this will help 

 

https://goo.gl/c9557m

Melody  Jul 15, 2017
 #2
avatar+90567 
+1

WolframAlpha tells me the answer is 1/4 but I'd really like someone to tell me why ://

Melody  Jul 16, 2017
 #3
avatar
0

Mathematica 11 gives the following as "Multiple-argument Formulas", but no step by step breakdown:

 

Cos[Pi/7] + Cos[Pi/7]^2 - 2 Cos[Pi/7]^3 == -1 + 2 Cos[Pi/14]^2 + (-1 + 2 Cos[Pi/14]^2)^2 - 2 (-1 + 2 Cos[Pi/14]^2)^3

 

Cos[Pi/7] + Cos[Pi/7]^2 - 2 Cos[Pi/7]^3 == 1 - 2 Sin[Pi/14]^2 + (1 - 2 Sin[Pi/14]^2)^2 - 2 (1 - 2 Sin[Pi/14]^2)^3

 

Cos[Pi/7] + Cos[Pi/7]^2 - 2 Cos[Pi/7]^3 == -3 Cos[Pi/21] + 4 Cos[Pi/21]^3 + (-3 Cos[Pi/21] + 4 Cos[Pi/21]^3)^2 - 2 (-3 Cos[Pi/21] + 4 Cos[Pi/21]^3)^3

Guest Jul 16, 2017
edited by Guest  Jul 16, 2017
edited by Guest  Jul 16, 2017
edited by Guest  Jul 16, 2017
edited by Guest  Jul 16, 2017
 #4
avatar+18612 
+2
Best Answer

cos(pi/7) + (cos(pi/7))^2 - 2(cos(pi/7))^3 = ?

step by step

 

\(\begin{array}{|l|rcll|} \hline \mathbf{1. \text{ Substitution } } \\ x = \frac{\pi}{7} && \cos(\frac{\pi}{7}) + \cos^2(\frac{\pi}{7}) - 2\cos^3(\frac{\pi}{7}) \\ &=& \cos(x) + \cos^2(x) - 2\cos^3(x) \\ \hline \end{array}\)

 

\(\mathbf{2. \cos^2(x) \text{ to } \cos(2x) \text{ and } \cos^3(x) \text{ to } \cos(3x) }\)

\(\begin{array}{|rcll|} \hline \text{Formula}: \\ \cos(2x) &=& 2\cos^2(x) - 1 \qquad \text{ or } \qquad \cos^2(x) = \frac12\Big(1+\cos(2x)\Big) \\ \cos(3x) &=& 4\cos^3(x) - 3\cos(x) \qquad \text{ or } \qquad \cos^3(x) = \frac14\Big(3\cos(x)+\cos(3x)\Big) \\ \\ && \cos(x) + \cos^2(x) - 2\cos^3(x) \\ &=& \cos(x) + \frac12\Big(1+\cos(2x)\Big) - \frac24\Big(3\cos(x)+\cos(3x)\Big) \\ &=& \frac12\Big( 2\cos(x) + 1 + \cos(2x)-3\cos(x) - \cos(3x) \Big) \\ &=& \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ \hline \end{array} \)

 

3. Change minus to plus

\(\begin{array}{|l|rclcl|} \hline \text{Formula}: \\ \cos(x) = -\cos(\pi-x) \\\\ & \cos(\frac{\pi}{7}) &=& -\cos(\pi-\frac{\pi}{7}) \\ & &=& -\cos(\frac67 \pi) \\ & \mathbf{\cos(x)} &\mathbf{=}&\mathbf{ -\cos(6x)} \\\\ & \cos(\frac{3\pi}{7}) &=& -\cos(\pi-\frac{3\pi}{7}) \\ & &=& -\cos(\frac47 \pi) \\ & \mathbf{\cos(3x) } &\mathbf{=}& \mathbf{-\cos(4x)} \\\\ \hline \end{array} \\ \begin{array}{|rcll|} \hline && \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ &=& \frac12\Big( 1+ \cos(6x)+\cos(2x)+\cos(4x) \Big) \\ &=& \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ \hline \end{array} \)

 

\(\begin{array}{|l|rcll|} \hline \mathbf{4. \text{ Substitution } } \\ y = 2x && \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ &=& \frac12\Big( 1+\cos(y)+\cos(2y)+ \cos(3y) \Big) \\ \hline \end{array} \)

 

5. complex numbers \(1+\cos(y)+\cos(2y)+ \cos(3y) = \ ?\)

\(\begin{array}{|rcll|} \hline C &=& 1+\cos(y)+\cos(2y)+ \cos(3y) \\ S &=& \qquad \sin(y)+\sin(2y)+ \sin(3y) \\ \mathbf{T} &\mathbf{=}& \mathbf{ C+i\cdot S } \\\\ T &=& 1+\cos(y)+\cos(2y)+ \cos(3y)+i\cdot \Big( \sin(y)+\sin(2y)+ \sin(3y) \Big)\\ &=& 1+\underbrace{\cos(y)+i\cdot \sin(y)}_{=e^{iy}} +\underbrace{\cos(2y)+i\cdot \sin(2y)}_{=e^{i2y}} +\underbrace{ \cos(3y)+i\cdot \sin(3y)}_{=e^{i3y}} \\ &=& 1+ e^{iy} + e^{i2y} + e^{i3y} \\ &=& 1+ e^{iy} + (e^{iy})^2 + (e^{iy})^3 \qquad \text{geometric series } r=e^{iy} \\ T &=& 1+ r + r^2 + r^3 \\ \hline T &=& 1+ r + r^2 + r^3 \\ r\cdot T &=& \qquad r + r^2 + r^3 + r^4 \\ \hline r\cdot T - T &=& r^4 - 1 \\ T\cdot (r-1) &=& r^4 - 1 \\ T &=& \frac{r^4 - 1} {r-1} \\ &=& \frac{(e^{iy})^4 - 1} {e^{iy}-1} \\ \mathbf{T} & \mathbf{=} & \mathbf{\frac{e^{i4y} - 1} {e^{iy}-1}} \\ \hline \end{array}\)

 

\(\begin{array}{|l|rclcl|} \hline \text{Formula}: \\ e^{i\phi} - 1 = 2i\cdot \sin(\frac{\phi}{2})\cdot e^{i\frac{\phi}{2}} \\\\ & \mathbf{T} & \mathbf{=} & \mathbf{\frac{e^{i4y} - 1} {e^{iy}-1}} \\ & &=& \frac{ 2i\cdot \sin(2y) \cdot e^{i2y} } { 2i\cdot \sin(\frac{y}{2})\cdot e^{ i\frac{y}{2} }} \qquad | \qquad y = 2x \\ & &=& \frac{ \sin(4x) \cdot e^{i4x} } { \sin(x)\cdot e^{ ix }} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot e^{i4x-ix} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot e^{i3x} \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot \Big( \cos(3x) + i\cdot \sin(3x) \Big) \\ & &=& \underbrace{\frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x)}_{=C} + i\cdot \underbrace{ \frac{ \sin(4x) } { \sin(x)} \cdot \sin(3x) }_{=S} \\\\ \hline \end{array}\)

 

6. Solution

\(\begin{array}{|rcll|} \hline & C &=& 1+\cos(y)+\cos(2y)+ \cos(3y) \qquad | \qquad y = 2x \\ & &=& 1+\cos(2x)+\cos(4x)+ \cos(6x) \\ & &=& \frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x) \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \text{Formula}: \\ \sin(x) = \sin(\pi-x) \\\\ & \cos(\frac{4\pi}{7}) &=& \sin(\pi-\frac{4\pi}{7}) \\ & &=& \sin(\frac37 \pi) \\ & \mathbf{\sin(4x)} &\mathbf{=}&\mathbf{ \sin(3x)} \\\\ & \cos(\frac{6\pi}{7}) &=& \sin(\pi-\frac{6\pi}{7}) \\ & &=& \sin(\frac17 \pi) \\ & \mathbf{\sin(6x)} &\mathbf{=}&\mathbf{ \sin(x)} \\\\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \cos(\frac{\pi}{7}) + \cos^2(\frac{\pi}{7}) - 2\cos^3(\frac{\pi}{7}) \quad & | \quad x=\frac{\pi}{7} \\ &=& \cos(x) + \cos^2(x) - 2\cos^3(x) \\ &=& \frac12\Big( 1- \cos(x)+\cos(2x)-\cos(3x) \Big) \\ &=& \frac12\Big( 1+\cos(2x)+\cos(4x)+ \cos(6x) \Big) \\ &=& \frac12\Big( \frac{ \sin(4x) } { \sin(x)} \cdot \cos(3x) \Big) \quad & | \quad \mathbf{\sin(4x)= \sin(3x)}\\ &=& \frac12\Big( \frac{ \sin(3x) } { \sin(x)} \cdot \cos(3x) \Big) \quad & | \quad \sin(3x)\cos(3x)=\frac{\sin(6x)}{2} \\ &=& \frac12\Big( \frac{ \sin(6x) } { 2 \sin(x)} \Big) \quad & | \quad \mathbf{\sin(6x)=\sin(x)} \\ &=& \frac12\Big( \frac{ \sin(x) } { 2 \sin(x)} \Big) \\ &=& \frac12\Big( \frac12 \Big) \\ &=& \mathbf{\frac14} \\ \hline \end{array}\)

 

 

laugh

heureka  Jul 18, 2017
 #5
avatar+76929 
+1

 

Wow....that's impressive, heureka!!!!....that might be one of the most difficult problems ever submitted to the forum.....!!!

 

It seemingly employed just about every facet of trig ( along with a few other things, too ) ...

 

 

cool cool cool

CPhill  Jul 18, 2017
edited by CPhill  Jul 19, 2017
 #6
avatar+18612 
+1

Thank you, CPhill

 

heureka

 

laughlaughlaugh

heureka  Jul 19, 2017

19 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details