need helphave to simplify and i don't know how (1+cos a-i*sin a)/(1-cos a+i*sina)

neela | Student

To simplify:

(1+cos a-isin a)/(1-cos a+i*sina).

= (1+cosa-isin a)(1-cos a-i*sina)/(1-cos a+i*sina))(1-cos a-i*sina).



={1-2isina-(sin^2a+cos^2a}/{2+2cosa}, as (cos^a+sin^2a = 1.

= -2isina/2(1+cosa).

= -isina/(1+cosa).

= -i(2sin(a/2))*cos(a/2)/{1+2cos^2(a/2)-1}, as cos2A = 2cos^2A-1.

= -2i{sin(a/2)}/2cos^2(a/2)

= -i tan(a/2).


(1+cos a-i*sin a)/(1-cos a+i*sina) = -2i*sina/(1+cosa) = -i*tan(a/2).

giorgiana1976 | Student

We'll re-write the sum of numerator:

1+cos a = 2 [cos (a/2)]^2

We'll re-write the sum of denominator:

1-cos a = 2 [sin (a/2)]^2

We'll re-write the ratio:

(1+cos a-i*sin a)/(1-cos a+i*sina) = {2 [cos (a/2)]^2-i*sin a}/{2 [sin (a/2)]^2+i*sina}

We'll factorize and we'll get:

2 cos (a/2)*[cos (a/2) - i*sin (a/2)]/2 sin (a/2)*[sin (a/2) + i*cos(a/2)]

We'll simplify and we'll get:

cot (a/2)*{[cos (a/2) - i*sin (a/2)]/[sin (a/2) + i*cos(a/2)]}

We'll multiply by the conjugate of denominator:

cot (a/2)*{[cos (a/2) - i*sin (a/2)]*[sin (a/2) - i*cos(a/2)]/{[sin (a/2)]^2 + [cos (a/2)]^2}}

But {[sin (a/2)]^2 + [cos (a/2)]^2} = 1(fundamental formula of trigonometry)

cot (a/2)*{[cos (a/2) - i*sin (a/2)]*[sin (a/2) - i*cos(a/2)]

cot (a/2)*{cos (a/2)*sin (a/2) - i*[cos(a/2)]^2 - i*[sin(a/2)]^2-cos (a/2)*sin (a/2)}

We'll eliminate like terms:

-i*cot (a/2)* {[sin (a/2)]^2 + [cos (a/2)]^2} = -i*cot (a/2)

The simplified ratio is -i*cot (a/2).

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