+0  
 
0
216
2
avatar

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

Guest Mar 15, 2017
 #1
avatar+12560 
0

Arcsin  -  .4382 =  334.01 degrees      205.98 degrees

ElectricPavlov  Mar 15, 2017
edited by ElectricPavlov  Mar 15, 2017
 #2
avatar+19653 
0

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

9 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.