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# Trigonometry - Tangent

+2
182
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+160

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
+23884
+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}$$

May 7, 2019
#2
+107018
+2

Very impressive Heureka

Melody  May 7, 2019
#4
+23884
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Thank you, Melody !

heureka  May 7, 2019
#3
+23884
+3

Trigonometry

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

May 7, 2019
#5
+160
+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
+23884
+1

Thank you, OldTimer !

heureka  May 8, 2019