# Trigonometric Identities

The question is asking what cos3θcosθ−sin3θsinθ is equivalent to, in the form of either sinnθ or cosnθ, and I'm also supposed to work out n. I've tried rewriting sin3θ as sin(2θ+θ) but I just go down a huge rabbit hole and somehow come out with a quartic.

Help??
Think about the identity
cos(A+B) = cosAcosB - sinAsinB

https://www.liverpool.ac.uk/~maryrees/homepagemath191/trigid.pdf
Original post by BobbJo
Think about the identity
cos(A+B) = cosAcosB - sinAsinB

https://www.liverpool.ac.uk/~maryrees/homepagemath191/trigid.pdf

okay, so I end up here:

Do I try to expand and simplify??
Original post by HollyR3
okay, so I end up here:

Do I try to expand and simplify??

no

look at cos3θcosθ sin3θsinθ

look at cos(A+B) = cosAcosB - sinAsinB
Original post by BobbJo
no

look at cos3θcosθ sin3θsinθ

look at cos(A+B) = cosAcosB - sinAsinB

oohhhhhhh I'm with you now, thanks!
Original post by Capossiali
oohhhhhhh I'm with you now, thanks!

But u didnt give the answer, im still stuck