+0  
 
0
666
2
avatar+157 

solve in trigonometric from
[2(cos 100°+sin 100°)] [5(cos300°+i sin 300°)]

 Apr 14, 2016
 #1
avatar+129907 
+5

[2(cos 100°+sin 100°)] [5(cos300°+i sin 300°)]   =

 

10 [ -sin (pi / 18) + i cos(pi/18)]  [ 1/2 - (√3/2) i]   [exact form]

 

10 [ -0.1736 + .9848i] [ .5 - . 866i]  =

 

10 [ -0.0868 + .4924i + 0.15033i - 0.8528i^2]  =

 

10 [ -0.0868 + 0.8528 + .64273i]  =

 

10 [ .766 + .64273i] =

 

7.66 + 6.4273i        [approximation]

 

 

 

cool cool cool

 Apr 14, 2016
 #2
avatar+157 
0

The answer given to this is 
10(cos 40°+i sin 40°)
smileyCan you please explain , thank you smiley

ariannasofia1  Apr 14, 2016

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