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i need the answer and solution preferably please. Thank you very much!

 Feb 22, 2019
 #2
avatar+23278 
+4

i need the answer and solution preferably please.

 

if  \(\large{ \tan(120^\circ-x)=\dfrac{\sin(120^\circ) -\sin(x)} {\cos(120^\circ) -\cos(x)} }\), where \(\large{0^\circ < x < 180^\circ}\), compute \(x\)

 

\(\text{Let $ \mathbf{s} = \sin(120^\circ)= \sin(60^\circ)=\dfrac{\sqrt{3}} {2} $} \\ \text{Let $ \mathbf{c} = \cos(120^\circ)=-\cos(60^\circ)=-\dfrac{1} {2} $} \\ \)

LHS:

\(\begin{array}{|rcll|} \hline \tan(120^\circ-x) &=& \dfrac{\sin(120^\circ-x)} {\cos(120^\circ-x)} \\\\ &=& \dfrac{\sin(120^\circ)\cos(x)-\cos(120^\circ)\sin(x)} {\cos(120^\circ)\cos(x)+\sin(120^\circ)\sin(x)} \\\\ &=& \dfrac{s\cdot \cos(x)-c\cdot\sin(x)} {c\cdot\cos(x)+s\cdot\sin(x)} \\ \hline \end{array}\)

 

RHS:

\(\begin{array}{|rcll|} \hline \dfrac{\sin(120^\circ) -\sin(x)} {\cos(120^\circ) -\cos(x)} \\\\ &=& \dfrac{ 2\cos\left(\dfrac{120^\circ+x}{2}\right) \sin\left(\dfrac{120^\circ-x}{2}\right) } {-2\sin\left(\dfrac{120^\circ+x}{2}\right) \sin\left(\dfrac{120^\circ-x}{2}\right)} \\\\ &=& -\dfrac{ \cos\left(\dfrac{120^\circ+x}{2}\right) } { \sin\left(\dfrac{120^\circ+x}{2}\right) } \\\\ &=& -\dfrac{ \cos\left(60^\circ+\dfrac{x}{2}\right) } { \sin\left(60^\circ+\dfrac{x}{2}\right) } \\\\ &=& -\dfrac{\cos(60^\circ)\cos\left(\dfrac{x}{2}\right) -\sin(60^\circ)\sin\left(\dfrac{x}{2}\right) } {\sin(60^\circ)\cos\left(\dfrac{x}{2}\right) +\cos(60^\circ)\sin\left(\dfrac{x}{2}\right) } \\\\ &=& \dfrac{\sin(60^\circ)\sin\left(\dfrac{x}{2}\right) -\cos(60^\circ)\cos\left(\dfrac{x}{2}\right) } {\sin(60^\circ)\cos\left(\dfrac{x}{2}\right) +\cos(60^\circ)\sin\left(\dfrac{x}{2}\right) } \\\\ &=& \dfrac{s\cdot \sin\left(\dfrac{x}{2}\right) +c\cdot \cos\left(\dfrac{x}{2}\right) } {s\cdot \cos\left(\dfrac{x}{2}\right) -c\cdot \sin\left(\dfrac{x}{2}\right) } \\ \hline \end{array}\)

 

\(\begin{array}{|lcll|} \hline \dfrac{s\cdot \cos(x)-c\cdot\sin(x)} {c\cdot\cos(x)+s\cdot\sin(x)} = \dfrac{s\cdot \sin\left(\dfrac{x}{2}\right) + c\cdot \cos\left(\dfrac{x}{2}\right) } {s\cdot \cos\left(\dfrac{x}{2}\right) - c\cdot \sin\left(\dfrac{x}{2}\right) } \\\\ (s\cdot \cos(x)-c\cdot\sin(x)) \left(s\cdot \cos\left(\dfrac{x}{2}\right) - c\cdot \sin\left(\dfrac{x}{2}\right) \right) \\ = (c\cdot\cos(x)+s\cdot\sin(x)) \left( s\cdot \sin\left(\dfrac{x}{2}\right) + c\cdot \cos\left(\dfrac{x}{2}\right) \right) \\\\ s^2\cos(x)\cos\left(\dfrac{x}{2}\right) - sc\cos(x)\sin\left(\dfrac{x}{2}\right)- sc\sin(x)\cos\left(\dfrac{x}{2}\right)+c^2\sin(x)\sin\left(\dfrac{x}{2}\right) \\ =c^2\cos(x)\cos\left(\dfrac{x}{2}\right) + sc\cos(x)\sin\left(\dfrac{x}{2}\right)+ sc\sin(x)\cos\left(\dfrac{x}{2}\right)+s^2\sin(x)\sin\left(\dfrac{x}{2}\right) \\\\ \cos(x)\cos\left(\dfrac{x}{2}\right)(s^2-c^2)-\sin(x)\sin\left(\dfrac{x}{2}\right)(s^2-c^2)-2sc\cdot \cos(x)\sin\left(\dfrac{x}{2}\right)-2sc\cdot \sin(x)\cos\left(\dfrac{x}{2}\right) = 0 \\\\ (s^2-c^2)\left(\underbrace{ \cos(x)\cos\left(\dfrac{x}{2}\right)-\sin(x)\sin\left(\dfrac{x}{2}\right) }_{=\cos\left(x+\dfrac{x}{2}\right)} \right) -2sc\cdot\left(\underbrace{ \cos(x)\sin\left(\dfrac{x}{2}\right)+\sin(x)\cos\left(\dfrac{x}{2}\right) }_{=\sin\left(x+\dfrac{x}{2}\right)} \right) = 0 \\\\ (s^2-c^2)\cos\left(x+\dfrac{x}{2}\right) - 2sc\cdot \sin\left(x+\dfrac{x}{2}\right) = 0 \\\\ 2sc\cdot \sin\left(x+\dfrac{x}{2}\right) = (s^2-c^2)\cos\left(x+\dfrac{x}{2}\right) \\\\ 2sc\cdot \sin\left(\dfrac{3x}{2}\right) = (s^2-c^2)\cos\left(\dfrac{3x}{2}\right) \\\\ \tan\left(\dfrac{3x}{2}\right) = \dfrac{s^2-c^2} {2sc} \quad | \quad s^2-c^2 = \dfrac34-\dfrac14 = \dfrac12 \\\\ \tan\left(\dfrac{3x}{2}\right) = \dfrac{\dfrac12} {2sc} \quad | \quad 2sc = 2\dfrac{\sqrt{3}}{2}\left(-\dfrac12\right)=-\dfrac{\sqrt{3}}{2} \\\\ \tan\left(\dfrac{3x}{2}\right) = \dfrac{\dfrac12} {-\dfrac{\sqrt{3}}{2}} \\\\ \tan\left(\dfrac{3x}{2}\right) = -\dfrac{\sqrt{3}}{3} \\\\ \dfrac{3x}{2} = \arctan(-\dfrac{\sqrt{3}}{3}) +n\cdot 180^\circ \qquad n\in \mathbb{Z} \\\\ x= \dfrac{2}{3}\cdot \arctan\left(-\dfrac{\sqrt{3}}{3}\right) +\dfrac{2}{3}\cdot n\cdot 180^\circ \\\\ x= \dfrac{2}{3}\cdot (-30^\circ) + n\cdot 120^\circ \\\\ x= -20^\circ + n\cdot 120^\circ \qquad n = 1 \quad (0^\circ < x < 180^\circ) \\\\ x= -20^\circ + 120^\circ \\\\ \mathbf{x= 100^\circ } \\ \hline \end{array}\)

 

laugh

 Feb 22, 2019
 #3
avatar+104704 
+2

Very impressive, Heureka !!!!!

 

 

cool cool cool

CPhill  Feb 22, 2019

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