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Thread: 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 $\displaystyle sin^2\theta$ for $\displaystyle -180\leq \theta \leq180$

    and the example shows these first few lines of working:

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

    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 $\displaystyle sin^2$

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

    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 $\displaystyle sin^2\theta$ for $\displaystyle -180\leq \theta \leq180$

    and the example shows these first few lines of working:

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

    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 $\displaystyle sin^2$

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

    thanx for your help in advance
    This is because,

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

    Also $\displaystyle \sin$ and $\displaystyle \theta$ are not separate. This whole $\displaystyle \sin \theta$ 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 $\displaystyle sin\theta=50$ then to find $\displaystyle \theta$ dont you divide it by sin or multiply it by $\displaystyle sin^{-1}$ making $\displaystyle \theta$ 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 $\displaystyle sin\theta=50$ then to find $\displaystyle \theta$ dont you divide it by sin or multiply it by $\displaystyle sin^{-1}$ making $\displaystyle \theta$ separate ?
    1. $\displaystyle sin\theta=50$ is impossible for real values of $\displaystyle \theta$.

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

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

    4. Things like $\displaystyle \sqrt{}, ~ \sin, ~ \sin^{-1}$ 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 $\displaystyle sin\theta=\frac{1}{2}$

    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 $\displaystyle sin^2\theta$ instead of $\displaystyle (sin\theta)^2$

    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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