Question

solve sin 2 theta

Answer 1

\(\bf sin(2\theta)\qquad ?\)

not much to solve, since there's no equation per se

Answer 2

\[\sin2\theta=\cos \theta \]

Answer 3

\(\bf {\color{blue}{ sin(2\theta)\implies 2sin(\theta)cos(\theta)}}
\\ \quad \\ \quad \\

sin(2\theta)=cos(\theta)\implies {\color{blue}{ 2sin(\theta)cos(\theta)}}=cos(\theta)\implies 2sin(\theta)=1
\\ \quad \\
sin(\theta)=\cfrac{1}{2}\implies sin^{-1}[sin(\theta)]=sin^{-1}\left(\cfrac{1}{2}\right)\implies \theta=sin^{-1}\left(\cfrac{1}{2}\right)\)

Answer 4

I still don't understand because I was thinking that I would have to use the unit circle to find the final answer

Answer 5

you do

Answer 6

so...\(\theta\) = "the angle whose sine is \(\cfrac{1}{2}\)"

Answer 7

Oh I see now so it would be \[\pi/6, 5\pi/6, \pi/2, 3\pi/2\]

Answer 8

yes, assuming you're meant to use a range between 0 and \(\pi\)

Answer 9

if you weren't given a specific range
then the inverse sine function restrictions will apply
that is the angle returned by the inverse sine function will be \(\bf \frac{\pi}{2}\le \theta\le-\frac{\pi}{2}\)

Answer 10

yes, assuming you're meant to use a range between 0 and \(2\pi\) I meant, anywayh

Answer 11

So it's ranging from [0,\[2\pi)\]

Answer 12

ok.. then your angles are correct :)

Answer 13

Thank you so much :)

Answer 14

yw