Asked by TP
Could someone help me with this question
Write in polar form : sinx - i cosx
I've come up to this point, and it's wrong, can't someone please help?
=sin(pie/2-x)-icos(pie/2-x)
=cosx-isinx
=cis(-x)
But the answer is cis(x-pie/2)
Write in polar form : sinx - i cosx
I've come up to this point, and it's wrong, can't someone please help?
=sin(pie/2-x)-icos(pie/2-x)
=cosx-isinx
=cis(-x)
But the answer is cis(x-pie/2)
Answers
Answered by
Reiny
almost
sinx = cos(pi/2 - x) and cosx = sin(pi/2 - x)
that is, the sine of any angle equals the cosine of its compliment.
so sinx - i cosx
= cos(pi/2 - x) - i sin(pi/2 - x)
= cis(pi/2 - x)
this webpage has some good examples with simple diagrams to show your topic
http://www.intmath.com/Complex-numbers/4_Polar-form.php
sinx = cos(pi/2 - x) and cosx = sin(pi/2 - x)
that is, the sine of any angle equals the cosine of its compliment.
so sinx - i cosx
= cos(pi/2 - x) - i sin(pi/2 - x)
= cis(pi/2 - x)
this webpage has some good examples with simple diagrams to show your topic
http://www.intmath.com/Complex-numbers/4_Polar-form.php
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