COMPLEX NUMBER mix TRIG (1 Viewer)

[ne plus ultra]

the question is shown as below:

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Express cos(6x) and sin(6x) in terms of cos(x) and sin(x) and hence tan(6x) in terms of tan(x).

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I even dont know how to start at all...><
Thanks!~

 

[conman]

Yep he's right using De Moivre's theorem to solve it!!!

 

[Shortbreads]

*does a dance for De Moivre*

Woo!

 

[ne plus ultra]

(cosx + isinx)⁶

Using de Moivre's
= cos6x + isin6x

Using the binomial theorem
= cos⁶x + 6isinxcos⁵x - 15sin²xcos⁴x - 20isin³xcos³x + 15sin⁴xcos²x + 6isin⁵xcosx - sin⁶x

Equate real coefficients
cos6x = cos⁶x - 15(1 - cos²x)cos⁴ + 15(1 - cos²x)²cos²x - (1 - cos²x)³)

Expand that, do likewise for sin6x and then put sin6x over cos6x. A tough and long process - usually in exams you'll be asked 3x or 4x

solved. thank you very much!

 

[ianc]

did he commit suicide?

 

[toadstooltown]

That would be cheating

 

[Shortbreads]

It would be kinda cutting off your own nose to spite your face though...

 

[Slidey]

Kinda. On that note, I've never seen this question turn up in an external exam.

Why? Because it is an easy process to follow, but it is also lengthy. That is to say: regardless of how 'complex' you deem it, it doesn't test component B.