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Math Help - Having trouble understanding this example

  1. #1
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    Having trouble understanding this example

    Hi my maths text book has a question that says:

    Solve sin^2\theta for -180\leq \theta \leq180

    and the example shows these first few lines of working:

    sin^2\theta=\frac{1}{2}
    sin\theta = \pm\sqrt{\frac{1}{2}}
    sin\theta =\pm\frac{1}{\sqrt{2}}

    but what iam wandering is what about the theta (angle) if you square root one side wouldn't you need to square root the theta as well not just the sin^2

    and wouldn't the second line look like this
    sin\sqrt{\theta}=\pm\sqrt{\frac{1}{2}}

    thanx for your help in advance
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  2. #2
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    Quote Originally Posted by silverbird View Post
    Hi my maths text book has a question that says:

    Solve sin^2\theta for -180\leq \theta \leq180

    and the example shows these first few lines of working:

    sin^2\theta=\frac{1}{2}
    sin\theta = \pm\sqrt{\frac{1}{2}}
    sin\theta =\pm\frac{1}{\sqrt{2}}

    but what iam wandering is what about the theta (angle) if you square root one side wouldn't you need to square root the theta as well not just the sin^2

    and wouldn't the second line look like this
    sin\sqrt{\theta}=\pm\sqrt{\frac{1}{2}}

    thanx for your help in advance
    This is because,

    (\sin \theta)^2 is written as \sin^2 \theta

    Also \sin and \theta are not separate. This whole \sin \theta is one term.
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  3. #3
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    that would explain it i had a feeling that might of been it. thx for your help
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  4. #4
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    Are you sure they are not separate i was just thinking what about if you have say sin\theta=50 then to find \theta dont you divide it by sin or multiply it by sin^{-1} making \theta separate ?
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  5. #5
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    Quote Originally Posted by silverbird View Post
    Are you sure they are not separate i was just thinking what about if you have say sin\theta=50 then to find \theta dont you divide it by sin or multiply it by sin^{-1} making \theta separate ?
    1. sin\theta=50 is impossible for real values of \theta.

    2. If I said that \sqrt{x} = 2 would you attempt to solve this by dividing both sides by \sqrt{} ?

    3. If you have \sin \theta = \frac{1}{\sqrt{2}} then one of the solutions is \theta = \sin^{-1} \frac{1}{\sqrt{2}}. (But don't even think of worrying about where the other solutions come from until you understand the basic concepts in this post). \sin^{-1} is an operator and ACTS on the number \frac{1}{\sqrt{2}}. It DOES NOT multiply the number.

    4. Things like \sqrt{}, ~ \sin, ~ \sin^{-1} are operators NOT numbers. They ACT on numbers. They are NOT treated like numbers.
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  6. #6
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    1. is impossible for real values of .
    sorry about that i was rushing and didn't think. I meant something like sin\theta=\frac{1}{2}

    is an operator and ACTS on the number . It DOES NOT multiply the number.
    Know everything makes sense i fully understand know. Its kind of weird how they use sin^2\theta instead of (sin\theta)^2

    Anyway i cant imagine how long it took you to right all that, thank you very much you explained everything great and you have given me a better understanding of maths (because i never thought of operators before).

    thank you mr fantastic
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