Question

4sin6α/cos^2 3α−sin^2 3α

Answer 1

Is that \[\frac{4sin6α}{cos^2 3α−sin^2 3α}\]?

Answer 2

yes

Answer 3

\[\cos^2\theta-\sin^2\theta=\cos(2\theta)\]

Answer 4

\[\frac{4sin6α}{cos^2 3α−sin^2 3α}=\frac{4sin6α}{cos 6α}=4tan 6a\]