Question

Help ?!?!?!?!

Answer 1

It's this right?\[\cos(2\theta) = 2\cos^2(\theta) - 1\]

Answer 2

No, the question.

Answer 3

but the first 2 and theta are not that close

Answer 4

lol, the same thing.

Answer 5

this is the picture..o lol

Answer 6

\[\textbf{HINT}\]\(\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)\)

Answer 7

It's just a hint. Can you now find an expression for \(\sin^2(\theta)\) in terms of \(\cos^2(\theta)\)?

Answer 8

no im not really good with this

Answer 9

Well, \(\sin^2\theta = 1 - \cos^2\theta\)

Answer 10

ok

Answer 11

Substitute \(1 - \cos^2\theta\) in \(\sin^2\theta\)

Answer 12

ok

Answer 13

i think the question ask "give an angle" to prove LHS=RHS, so
we can use an angle for example the 0 degrees, pi/3, pi/4, or the others .
maybe.. :)

Answer 14

ok...so what should i writes as my answer...im confused

Answer 15

u can use 0 deg, and then plug it into ur question above to prove is LHS=RHS or no

Answer 16

ok