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??
Help??
Think about the identity
cos(A+B) = cosAcosB - sinAsinB
https://www.liverpool.ac.uk/~maryrees/homepagemath191/trigid.pdf
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
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??
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
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
Quick Reply
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