We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+2
49
6
avatar+124 

Hi Tutors...hope you are having a great day! I managed to find values for delta but could not show solution to equation.

Regards and thanks.

 May 7, 2019
 #1
avatar+22180 
+4

Trigonometry

 

\(\begin{array}{|rcll|} \hline \tan{\theta} &=& \lambda\ \tan(A-\theta) \\ \boxed{\tan{\theta}=\dfrac{\sin(\theta)}{\cos(\theta)} \\ \tan(A-\theta) = \dfrac{\sin(A-\theta)}{\cos(A-\theta)} } \\ \dfrac{\sin(\theta)}{\cos(\theta)} &=& \lambda\ \left(\dfrac{\sin(A-\theta)}{\cos(A-\theta)} \right) \\ \boxed{ \sin(A-\theta) = \sin(A)\cos(\theta)-\cos(A)\sin(\theta) \\ \cos(A-\theta) = \cos(A)\cos(\theta)+\sin(A)\sin(\theta) } \\ \dfrac{\sin(\theta)}{\cos(\theta)} &=& \lambda\ \left(\dfrac{\sin(A)\cos(\theta)-\cos(A)\sin(\theta)} {\cos(A)\cos(\theta)+\sin(A)\sin(\theta)} \right) \\ \sin(\theta) \Big(\cos(A)\cos(\theta)+\sin(A)\sin(\theta) \Big) &=& \lambda\ \cos(\theta) \Big(\sin(A)\cos(\theta)-\cos(A)\sin(\theta) \Big) \\ \cos(A)\sin(\theta)\cos(\theta)+ \sin(A)\sin^2(\theta) &=& \lambda\ \cos^2(\theta)\sin(A)-\lambda\ \cos(A)\sin(\theta)\cos(\theta) \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta) &=& \lambda\ \cos^2(\theta)\sin(A)-\sin(A)\sin^2(\theta) \\ \boxed{ \cos^2(\theta) = 1-\sin^2(\theta) } \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta) &=& \lambda\ \Big(1-\sin^2(\theta) \Big) \sin(A)-\sin(A)\sin^2(\theta) \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta) &=& \lambda\ \sin(A)-(\lambda+1) \sin(A)\sin^2(\theta) \\ (\lambda+1)\cos(A)\sin(\theta)\cos(\theta)+(\lambda+1) \sin(A)\sin^2(\theta) &=& \lambda\ \sin(A) \\ (\lambda+1)\Big( \cos(A)\sin(\theta)\cos(\theta)+\sin(A)\sin^2(\theta) \Big) &=& \lambda\ \sin(A) \\ \boxed{\sin(\theta)\cos(\theta)=\dfrac{\sin(2\theta)}{2} \\ \sin^2(\theta) = \dfrac{1-\cos(2\theta)}{2} } \\ (\lambda+1)\Big( \cos(A)\dfrac{\sin(2\theta)}{2}+\sin(A) \Big(\dfrac{1-\cos(2\theta)}{2}\Big) \Big) &=& \lambda\ \sin(A) \\ (\lambda+1)\Big( \cos(A)\sin(2\theta)+\sin(A) \Big( 1-\cos(2\theta)\Big) \Big) &=& 2\lambda\ \sin(A) \\ (\lambda+1)\Big( \cos(A)\sin(2\theta)-\sin(A)\cos(2\theta) + \sin(A) \Big) &=& 2\lambda\ \sin(A) \\ \boxed{ \cos(A)\sin(2\theta)-\sin(A)\cos(2\theta) = \sin(2\theta-A) } \\ (\lambda+1)\Big( \sin(2\theta-A) + \sin(A) \Big) &=& 2\lambda\ \sin(A) \\ (\lambda+1)\sin(2\theta-A) +(\lambda+1)\sin(A) &=& 2\lambda\ \sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& 2\lambda\ \sin(A)-(\lambda+1)\sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& 2\lambda\ \sin(A)-\lambda\ \sin(A)- \sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& \lambda\ \sin(A)- \sin(A) \\ (\lambda+1)\sin(2\theta-A) &=& (\lambda -1) \sin(A)\\ \mathbf{(\lambda -1) \sin(A)} &=& \mathbf{(\lambda+1)\sin(2\theta-A)} \\ \hline \end{array}\)

 

laugh

 May 7, 2019
 #2
avatar+100811 
+2

Very impressive Heureka    laugh

Melody  May 7, 2019
 #4
avatar+22180 
+3

Thank you, Melody !

 

laugh

heureka  May 7, 2019
 #3
avatar+22180 
+3

 Trigonometry

 

\(\begin{array}{|rcll|} \hline \mathbf{\tan{\theta}} &=& \mathbf{2\ \tan(60^\circ-\theta)} \quad | \quad \lambda = 2,\ A=60^\circ \\\\ (2-1)\sin{60^\circ} &=& (2+1)\sin(2\theta-60^\circ) \\ \sin{60^\circ} &=& 3\sin(2\theta-60^\circ) \quad | \quad \sin(60^\circ)=\dfrac{\sqrt{3} }{2} \\ \dfrac{\sqrt{3} }{2} &=& 3\sin(2\theta-60^\circ) \\ \mathbf{ \sin(2\theta-60^\circ)} &=& \mathbf{\dfrac{\sqrt{3}}{6}} \\ 2\theta-60^\circ &=& \arcsin\left(\dfrac{\sqrt{3}}{6}\right) +360^\circ n,\ n\in\mathbb{Z} \\ 2\theta &=& 60^\circ +16.7786548810^\circ +360^\circ n \\ 2\theta &=& 76.7786548810^\circ +360^\circ n \\ \theta &=& 38.3893274405^\circ +180^\circ n \\ \\ \mathbf{\theta_1} &=& \mathbf{38.3893274405^\circ} \\\\ \theta_2 &=& 38.3893274405^\circ+180^\circ \\ \mathbf{\theta_2} &=& \mathbf{218.389327440^\circ} \\ \hline \sin(2\theta-60^\circ) &=& \sin\Big(180^\circ-(2\theta-60^\circ)\Big) \\ &=& \sin(180^\circ-2\theta+60^\circ) \\ &=& \sin(240^\circ-2\theta ) \\\\ \mathbf{ \sin(240^\circ-2\theta )} &=& \mathbf{ \dfrac{\sqrt{3}}{6} } \\ 240^\circ-2\theta &=& \arcsin\left(\dfrac{\sqrt{3}}{6}\right) +360^\circ n,\ n\in\mathbb{Z} \\ 2\theta &=& 240^\circ-16.7786548810^\circ - 360^\circ n \\ 2\theta &=& 223.221345119^\circ - 360^\circ n \\ \theta &=& 111.610672560^\circ - 180^\circ n \\ \\ \mathbf{\theta_3} &=& \mathbf{111.610672560^\circ} \\\\ \theta_4 &=& 111.610672560^\circ+180^\circ \\ \mathbf{\theta_4} &=& \mathbf{291.610672560^\circ} \\ \hline \end{array}\)

 

 

laugh

 May 7, 2019
 #5
avatar+124 
+1

Great job Heureka! and mostly thanks for your clarifications (boxed) to support the solution flow! As they say in Italy you know one more than the devil!

 May 8, 2019
 #6
avatar+22180 
+1

Thank you, OldTimer !

 

laugh

heureka  May 8, 2019

9 Online Users

avatar