vomplex numbre (1 Viewer)

[kractus]

Not sure where to start for part b. I tried it a few times starting with RHS and got nowhere and I'm not sure how to introduce cos^3 and sin^4 for LHS

 

[chilli 412]

vomplex numbre

 

[Masaken]

View attachment 37420
Not sure where to start for part b. I tried it a few times starting with RHS and got nowhere and I'm not sure how to introduce cos^3 and sin^4 for LHS

start from LHS = 128cos^3sin^4 (pretend the theta is there, and don't bother with starting with RHS, this way is quicker).
then split it up, since you're aiming to get a form similar to z^n + (or -) 1/z^n or 2cosn or 2isinn (which helps you fulfil the term 'Hence')
you will get 2^3 x 2^4 x cos^3 x sin^4 (again, pretend theta is there), which can then be simplified by grouping the common powers together to get (2cos)^3 x (2sin)^4. using the identities from part (a), you can quickly equate them to be (z + 1/z)^3 x (z-1/z)^4. expand each of the individual brackets out, then you'll end up getting the RHS.

 

[tywebb]

Cambridge again?

Maybe this

 

[Etho_x]

Excuse the messy iPad writing (I’m in bed lol) but here’s the process of the expansion above

Edit: that should be a minus where I wrote z + 1/z to 4th power, my brain not working zzz