Question

Help? Please?

Answer 1

note that: \(\sec \theta = \frac{1}{\cos \theta}\)
and,\(\sin \theta = \sqrt{1-\cos^2 \theta}\)
are you familiar with the double angle formulas?

Answer 2

is there a way I can get the entire solved problem with work? it's not for me it's for a friend who is struggling. Please?

Answer 3

lol i think it's better if you can do it, or i can do a step-by step tutorial

Answer 4

ummm how can I say this, I have never done this stuff so I am an idiot with math, and my friend is busy doing 2 or 3 other subjects at once.

Answer 5

i might be able to do it step by step

Answer 6

okay, first of all, \(\sin 2\theta = 2 \sin \theta \cos \theta\)
and \(\cos \theta = \frac{2}{3}\) from the relation above. use the formula above to get sine and sub it in. you follow?

Answer 7

*embarrassed* sorry... :( no

Answer 8

sorry, i'll adjust my solution.
\(\sin 2 \theta = 2\sin \theta \cos \theta\)
\(\sec \theta = \frac{1}{\cos \theta}\)
so, \(\cos\theta=\frac{1}{\sec \theta}=\frac{1}{3/2}=2/3\)
\(\sin\theta =\sqrt{1-\cos^2\theta}=\sqrt{1-(2/3)^2}=\sqrt5/3\)
sub it all into the first formula, \(\sin 2 \theta = 2 (\sqrt5/3)(2/3)\)

Answer 9

O.O..... thank you for trying it's much appreciated.

Answer 10

lol those are formulas. you might want to review(or ask your friend review them)