+0

​geometry proof

0
94
2

Feb 13, 2020

#1
+2850
0

Ughhh so hard! I tried using law of cosines, but it didn't work! I got a massive equation that I didn't know how to simplify!

Feb 14, 2020
#2
+24430
+2

​geometry proof

$$\begin{array}{|rcll|} \hline \mathbf{\text{In ACP sin-rule:}} \\ \hline \frac{w}{\sin(90^\circ-\alpha)} &=& \frac{CP}{\sin(45^\circ)} \quad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{In CPQ sin-rule:}} \\ \hline \frac{\sin(45^\circ)}{v} &=& \frac{\sin(\alpha)}{CP} \quad (2) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)\times(2): & \frac{w}{\sin(90^\circ-\alpha)}\times \frac{\sin(45^\circ)}{v} &=&\frac{CP}{\sin(45^\circ)} \times \frac{\sin(\alpha)}{CP} \\\\ & \frac{w}{\cos(\alpha)}\times \frac{\sin(45^\circ)}{v} &=&\frac{\sin(\alpha)}{\sin(45^\circ)} \\\\ & \frac{w}{v} &=& \frac{\sin(\alpha)\sin(\alpha)} {\sin^2(45^\circ)} \\\\ & \frac{w}{v} &=& \frac{2\sin(\alpha)\sin(\alpha)} {2\sin^2(45^\circ)} \\ & && \boxed{\sin(45^\circ) = \frac{\sqrt{2}}{2} \\ \sin^2(45^\circ) = \frac{1}{2} }\\ & \frac{w}{v} &=& \frac{2\sin(\alpha)\sin(\alpha)} {2*\frac{1}{2}} \\ & \frac{w}{v} &=& 2\sin(\alpha)\sin(\alpha) \\ & && \boxed{ 2\sin(\alpha)\sin(\alpha)= \sin(2\alpha) }\\ & \mathbf{\frac{w}{v}} &=& \mathbf{\sin(2\alpha)} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\text{In BCQ sin-rule:}} \\ \hline \frac{u}{\sin(\alpha-45^\circ)} &=& \frac{CQ}{\sin(45^\circ)} \quad (3) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{In CPQ sin-rule:}} \\ \hline \frac{\sin(45^\circ)}{v} &=& \frac{\sin(135^\circ-\alpha)}{CQ} \quad (4) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (3)\times(4): & \frac{u}{\sin(\alpha-45^\circ)}\times \frac{\sin(45^\circ)}{v} &=&\frac{CQ}{\sin(45^\circ)} \times \frac{\sin(135^\circ-\alpha)}{CQ} \\\\ & \frac{u}{\sin(\alpha-45^\circ)}\times \frac{\sin(45^\circ)}{v} &=&\frac{\sin(135^\circ-\alpha)}{\sin(45^\circ)} \\\\ & \frac{u}{v} &=& \frac{\sin(\alpha-45^\circ)\sin(135^\circ-\alpha)} {\sin^2(45^\circ)} \\ & && \boxed{\sin(45^\circ) = \frac{\sqrt{2}}{2} \\ \sin^2(45^\circ) = \frac{1}{2} }\\ & \frac{u}{v} &=& \frac{\sin(\alpha-45^\circ)\sin(135^\circ-\alpha)} {\frac{1}{2}} \\\\ & \frac{u}{v} &=& 2\sin(\alpha-45^\circ)\sin(135^\circ-\alpha) \\ & && \boxed{\sin(135^\circ-\alpha) = \sin \Big(180^\circ-(135^\circ-\alpha) \Big)\\ = \sin(\alpha+45^\circ) }\\ & \frac{u}{v} &=& 2\sin(\alpha-45^\circ)\sin(\alpha+45^\circ) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \mathbf{ \sin(\alpha-45^\circ)\sin(\alpha+45^\circ) } \\ &=&\Big(\sin(\alpha)\cos(45^\circ)-\cos(\alpha)\sin(45^\circ)\Big) \Big(\sin(\alpha)\cos(45^\circ)+\cos(\alpha)\sin(45^\circ)\Big) \\ && \boxed{ \cos(45^\circ)=\sin(45^\circ)=\frac{\sqrt{2}}{2} } \\ &=&\frac{\sqrt{2}}{2}\Big(\sin(\alpha)-\cos(\alpha)\Big) \frac{\sqrt{2}}{2}\Big(\sin(\alpha)+\cos(\alpha)\Big) \\ &=&\frac{1}{2}\Big(\sin(\alpha)-\cos(\alpha)\Big)\Big(\sin(\alpha)+\cos(\alpha)\Big) \\ &=&\frac{1}{2}\Big(\sin^2(\alpha)-\cos^2(\alpha)\Big) \\ &=&-\frac{1}{2}\Big(\cos^2(\alpha)-\sin^2(\alpha)\Big) \\ && \boxed{ \cos^2(\alpha)-\sin^2(\alpha)= \cos(2\alpha)} \\ &=&\mathbf{-\frac{1}{2}\cos(2\alpha)} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \frac{u}{v} &=& 2\sin(\alpha-45^\circ)\sin(\alpha+45^\circ) \\ \frac{u}{v} &=& 2\left(-\frac{1}{2}\cos(2\alpha)\right) \\ \mathbf{\frac{u}{v}} &=& \mathbf{-\cos(2\alpha)} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\frac{w}{v}} = \mathbf{\sin(2\alpha)} &\qquad & \mathbf{\frac{u}{v}} = \mathbf{-\cos(2\alpha)} \\\\ \left( \frac{w}{v} \right)^2 + \left( \frac{u}{v} \right)^2 &=& \Big( \sin(2\alpha)\Big)^2 + \Big(-\cos(2\alpha)\Big)^2 \\ \left( \frac{w}{v} \right)^2 + \left( \frac{u}{v} \right)^2 &=& \sin^2(2\alpha) + \cos^2(2\alpha) \\ \frac{w^2}{v^2} + \frac{u^2}{v^2} &=& 1 \quad | \quad \times v^2 \\ \mathbf{w^2 + u^2} &=& \mathbf{v^2} \\ \hline \end{array}$$

Feb 14, 2020