Need Help - De Moivre's Theorem (1 Viewer)

[blackops23]

Hi here's a question that I just can't do:

Q. Simplify: {[1 + sin pi/7 + icos pi/7]/[1+sin pi/7 - icos pi/7]}^7

I can't seem to see any trig rule that i can use to simplify this, help is greatly appreciated.

Thanks guys

 

[hscishard]

Double angle formula for cos to get rid of the 1. Double angle formula for sin to use the de moivre theorem. I remember doing this in the Terry Lee book

 

[blackops23]

oh damn, i forgot to put in the i, so basically unlike the previous terry lee questions, instead of the sine being imaginary, the cos is the imaginary part,

so i've no idea how to simplify....
i tried factorising the i to get it into cos(a) + isin(a), but nothing happened

please, i really need help

any hints or tips on what rules i must use to simplify this would really be nice
thanks guys

 

[Laplace3]

no one help him, he needs to get a tutor

 

[nohelp]

Use the fact that sin(a)=cos(pi/2-a) and vice versa. i.e cos(a)=sin(pi/2-a)
EDIT:Go with Rawrence's answer as the method I and hscishard posted will get you no where (just tried it)

 

[rawrence]

 

[blackops23]

thanks a lot mate! =D

but just a side question, this may sound pretty stupid - a basic 2 unit question, but what are you supposed to be looking for in order to factorise the numerator in STEP 2 into the numerator of STEP3 so quickly? I see you can factor the 2 out when after you convert the (1 + sin^2 pi/7 - cos^2 pi/7 --> 2sin^2 pi/7, but how do you go about factorising the rest?

Thanks

 

[rawrence]

I cut out steps so coding in latex wouldn't be so tedious, err this is a bit hard to explain because i cbf latexing. Can you just work backwards from where i factorised? I'm a bit busy

 

[xV1P3R]

Do you mean

 

[blackops23]

Do you mean

yes thankyou goawd that was a dumb question...

 

[laplace4]

man you are dumb