Thread: 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.

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

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

Originally Posted by JoshHJ

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

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

Thanks, but the answer I am given for the problem is: "45°". We haven't done arcsin yet.

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.

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

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

and

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

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