Three cube roots of 125(cos37°+ i sin37°) are ? answers in Trigonometric form please
Why dont you try it the satellite73 way? Commo'n You start and I will help you wherever possible
lol, i did ^^ and I'm stuck at the 3rd part, i got the other two answers though ^^
One answer is z=5(cos 12.33 + i sin 12.33) ..... You must have got this
ok answer 1 : 5(cos12.3°+ i sin12.3°) answer 2 : 5(cos132.3°+ i sin132.3°) answer 3: stuck -.-
yes yes !! ^^
glad i got something right lol :)
third answer = cos ( (12.3+ 720)/3) + i sin((12.3+720)/3)
let @satellite73 check this
he's off but your answer is good though ^^
Btw, your secind answer should be cos ( (12.3+ 360)/3) + i sin((12.3+360)/3)
you know what i think you're right
yea and it's 132.3 just like what i had :D
i think you're right with the third answer, do you just add another 360 to that 360 to get 720 ?
:D
and what about if the question is ask for a fourth root, then you add another 360 ?
Yes. But you have to divide by 4
For all roots
yea i understand that, but if it was fourth root, then what do you add on the fourth one? i mean on the third root u add it with 12.3 right? what about fourth root? do you still add it with 12.3 ? or 132.3 ?
Well for fourth root, it would be sqrt(5){ cos (37/4)+ i sin(37/4)}
2nd answer: sqrt(5){ cos (37+360/4)+ i sin(37+360/4)}
4th answer sqrt(5){ cos (37+1080/4)+ i sin(37+1080/4)}
ok i get it ^^
i kind did it different but as long as we got same answer then we'll good ^^
OK :D It 3 am here. I better catch some sleep
omg really ?
sorry for keeping you up so late :(
i didn't know lol -.-
No It's my pleasure learning math :)
ah :D, thanks again for your help ^^
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