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# Maths Compund Angles Trignometry

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

Jul 15, 2017

#4
+21848
+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

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

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

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

Jul 18, 2017

#1
+99331
+1

Maybe this will help

https://goo.gl/c9557m

Jul 15, 2017
#2
+99331
+1

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

Jul 16, 2017
#3
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

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
+21848
+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

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

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

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

heureka Jul 18, 2017
#5
+98172
+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 ) ...

Jul 18, 2017
edited by CPhill  Jul 19, 2017
#6
+21848
+1

Thank you, CPhill

heureka

heureka  Jul 19, 2017