Having trouble understanding this example

**silverbird** · March 20th 2009, 08:40 PM

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

**Shyam** · March 20th 2009, 08:47 PM

Originally Posted by silverbird

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.

**silverbird** · March 20th 2009, 09:43 PM

that would explain it i had a feeling that might of been it. thx for your help

**silverbird** · March 20th 2009, 09:52 PM

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 ?

**mr fantastic** · March 20th 2009, 10:02 PM

Originally Posted by silverbird

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.

**silverbird** · March 20th 2009, 11:33 PM

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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