Question 61673: sir,
i am doing B.Sc(maths) 1st year at madurai kamaraj university through distance education. when i refered 11th books i couldn't understand some steps.
please explain me.
represent the complex number in the polar form
let z= -square root of 3+i
here a=-square root of 3,b=1
r=2
cos theta=-square root of 3/ 2
sin theta=1/2
tan theta=-1/square root of 3
tan theta=-tan(pi/6)(till this i understood)
here they have given theta =150(how this has came) explain me
the same doubt in another problem
the question is write the complex number 1-square root of 3(i)
r=2
cos theta=1/2
sintheta=-square root of 3/2
tan theta=-1/square root of 3
tan theta=tan(-1/square root of 3)
here they have given theta =-60 (explain me)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! let z= -square root of 3+i
here a=-square root of 3,b=1
r=2
cos theta=-square root of 3/ 2
sin theta=1/2
tan theta=-1/square root of 3
tan theta=-tan(pi/6)(till this i understood)
here they have given theta =150(how this has came) explain me
----------
The sine is positive and the cosine is negative.
The reference angle is 30 degrees, but it is in
the 2nd quadrant, so theta = 150 degrees.
---------------
cos theta=1/2
sintheta=-square root of 3/2
tan theta=-1/square root of 3
tan theta=tan(-1/square root of 3)
here they have given theta =-60 (explain me)
-------
The sine is negative and the cosine is positive.
The reference angle is 60 degrees, but it is in
the 4th quadrant, so theta = 300 degrees or -60 degrees.
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Cheers,
Stan H.
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