Question

1-8sin^2beta(cos^2beta)

Answer 1

Hey Jam,
\(\Large\bf \color{#CC0033}{\text{Welcome to OpenStudy! :)}}\)

Answer 2

What is your problem? Need to simplify?

Answer 3

Do you mean \(1-8\sin^2\beta\cos^2\beta\) or \((1-8\sin^2\beta)\cos^2\beta\)?

Answer 4

we don't know what your problem is.
\[1-8\sin ^2\beta \cos ^2\beta =1-2(2\sin \beta \cos \beta)^2=1-2\sin ^2(2\beta)=\cos 4\beta \]

Answer 5

The first one, I need to establish the identity. I am trying to figure out how it ends up being cos(4beta), but am having some problems.

Answer 6

Thank you, I see it now.
On \[\sin(2\beta)+\sin(4\beta)/\cos(2\beta)+\cos(4\beta) \], how does it go from that to \[2\sin(3\beta)\cos(-\beta)/2\cos(3\beta)\cos(-\beta) \]?
I do not understand what identity they are using to do this. Perhaps the Even/Odd Properties?

Answer 7

\[\sin C+\sin D=2\sin \frac{ C+D }{2 }\cos \frac{ C-D }{2 }\]
\[\cos C+\cos D=2\cos \frac{ C+D }{ 2 }\cos \frac{ C-D }{ 2 }\]

Answer 8

Got it, Tan(3beta). <3 !

Answer 9

\[\sin 3x=3\sin x-4\sin ^3x,\cos 3x=4\cos ^3x-\cos x\]