# Solving trig equations- Stuck at the double angle forumla

• Feb 20th 2009, 07:59 PM
coasterguy
Solving trig equations- Stuck at the double angle forumla
I find it hard to type the problem on the computer, so I included a picture with a note on where I am stuck. I know I can't divide "1" by "sin2theta" to get 30 degrees, so I don't know what to do next. Thanks.

http://img.photobucket.com/albums/v6...1/P1010071.jpg
• Feb 20th 2009, 08:15 PM
JoshHJ
$
sin (2\theta) = 1
$

$
\arcsin (1) = 2\theta
$

$
\frac {\arcsin (1)}{2} = \theta
$

The sine function is 1 at 90 degrees or $\frac {\pi}{2}$
Therefore, $\theta = \frac {\pi}{2}$
• Feb 20th 2009, 09:04 PM
coasterguy
Quote:

Originally Posted by JoshHJ
$
sin (2\theta) = 1
$

$
\arcsin (1) = 2\theta
$

$
\frac {\arcsin (1)}{2} = \theta
$

The sine function is 1 at 90 degrees or $\frac {\pi}{2}$
Therefore, $\theta = \frac {\pi}{2}$

Thanks, but the answer I am given for the problem is: "45°". We haven't done arcsin yet.
• Feb 20th 2009, 09:22 PM
JoshHJ
Oops, the answer is sin 45. I meant to type pi/4. Though, if you are doing trig identities you should have gone over arcsin, which is the inverse function of sine.
• Feb 20th 2009, 09:24 PM
Reckoner
Quote:

Originally Posted by coasterguy
I find it hard to type the problem on the computer, so I included a picture with a note on where I am stuck. I know I can't divide "1" by "sin2theta" to get 30 degrees, so I don't know what to do next. Thanks.

As you have written, $0^\circ\leq2\theta<720^\circ\text.$ Where, in the interval $0^\circ\leq\phi<720^\circ$, is $\sin\phi$ equal to 1? That would be $\phi = 90^\circ, 450^\circ\text.$ But in our case, the angle is $\phi = 2\theta\text.$ So

$2\theta = 90^\circ\Rightarrow\theta = 45^\circ$

and

$2\theta = 450^\circ\Rightarrow\theta = 225^\circ\text.$

If you substitute these back into the original equation, you'll see that the solution is $45^\circ,$ and the second angle of $225^\circ$ is an extraneous solution introduced when you squared both sides.