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If sinA=-.4382. 0<A<360. Find two possible values for A.

Guest Mar 15, 2017
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 #1
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Arcsin  -  .4382 =  334.01 degrees      205.98 degrees

ElectricPavlov  Mar 15, 2017
edited by ElectricPavlov  Mar 15, 2017
 #2
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If sin(A) = -0.4382.  0 < A < 360.

Find two possible values for A.

 

\(\begin{array}{|rcll|} \hline \sin(A) &=& -0.4382 \\ A_1 &=& \arcsin(-0.4382) + z\cdot 360^{\circ} \quad & | \quad z \in \mathbb{Z} \\ A_1 &=& -25.9890903445^{\circ} + z\cdot 360^{\circ} \\ A_1 &=& -25.9890903445^{\circ}+360^{\circ} + z\cdot 360^{\circ} \\ \mathbf{A_1} &\mathbf{=}& \mathbf{334.010909656^{\circ} + z\cdot 360^{\circ}} \\\\ \sin(A)=\sin(180^{\circ}-A) &=&-0.4382 \\ 180^{\circ}-A_2 &=& \arcsin(-0.4382) + z\cdot 360^{\circ} \quad & | \quad z \in Z \\ A_2 &=& 180^{\circ}- \arcsin(-0.4382) + z\cdot 360^{\circ} \\ A_2 &=& 180^{\circ}- (-25.9890903445^{\circ}) + z\cdot 360^{\circ} \\ A_2 &=& 180^{\circ}+25.9890903445^{\circ}) + z\cdot 360^{\circ} \\ \mathbf{A_2} &\mathbf{=}&\mathbf{205.989090344^{\circ}+ z\cdot 360^{\circ}} \\ \hline \end{array} \)

 

0 < A < 360.

\(\begin{array}{|rcll|} \hline \mathbf{A_1} &\mathbf{=}& \mathbf{334.010909656^{\circ} } \\ \mathbf{A_2} &\mathbf{=}& \mathbf{205.989090344^{\circ}} \\ \hline \end{array}\)

 

laugh

heureka  Mar 15, 2017

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