Question

Guys please help :0

Answer 1

second part only
http://screencast.com/t/63QoGR6gM

Answer 2

@Hero @ParthKohli @amistre64

Answer 3

@AccessDenied @Preetha @Vincent-Lyon.Fr @radar

Answer 4

@eliassaab @ganeshie8

Answer 5

why just guys?

Answer 6

what do u mean?

Answer 7

you asked guys please help? I vas just wandering why just guys.

Answer 8

alright anyone can help, but i see no one is even trying to check out the question :(

Answer 9

what ist the question?

Answer 10

http://screencast.com/t/63QoGR6gM

Answer 11

QUESTION: Prove that the real part of \(\dfrac1{2\cos\theta+i(1-2\sin\theta)+2-i}\) is constant when \(-\pi<\theta<\pi\)

Answer 12

\[\frac1{2\cos\theta+i(1-2\sin\theta)+2-i}\]\[=\frac1{(2\cos\theta+2)-2i\sin\theta}\]\[=\frac{2\cos\theta+2+4i\sin\theta}{(2\cos\theta+2)^2+4\sin^2\theta}\]

Answer 13

Real part = \(\dfrac{2\cos\theta+2}{(2\cos\theta+2)^2+4\sin^2\theta}\)\[=\frac{2\cos\theta+2}{4\cos^2\theta+8\cos\theta+4+4\sin^2\theta}\]\[=\frac{2\cos\theta+2}{8\cos\theta+8}\]\[=\frac14\]

Answer 14

\[\therefore\mbox{The real part of the fraction in question is a constant.}\]

Answer 15

Thanks alot bro...I got it :)

Answer 16

Welcome :D