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Math Help - Solving trig equations- Stuck at the double angle forumla

  1. #1
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    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.

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  2. #2
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    <br />
sin (2\theta) = 1<br />
    <br />
\arcsin (1) = 2\theta<br />
    <br />
\frac {\arcsin (1)}{2} = \theta<br />

    The sine function is 1 at 90 degrees or \frac {\pi}{2}
    Therefore, \theta = \frac {\pi}{2}
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  3. #3
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    Quote Originally Posted by JoshHJ View Post
    <br />
sin (2\theta) = 1<br />
    <br />
\arcsin (1) = 2\theta<br />
    <br />
\frac {\arcsin (1)}{2} = \theta<br />

    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.
    Last edited by coasterguy; February 20th 2009 at 09:19 PM.
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  4. #4
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    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.
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    Quote Originally Posted by coasterguy View Post
    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.
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