Forum ► Technical Corner ► Math Problem Thread

Not really a problem but just a question.

Explain how Sine^{-1} turns sine into an angle, usually theta :P

Explain how Sine

4431.207 days ago

Apr 12, 2011 - 10:35 PM

Arcsine undoes Sine. The information you gather from a triangle is the Sine of ABC, which is not useful in finding the angle measurement. We can undo Sine though with Arcsine, just like how the sqrt(x^{2}) == x.

At least, I think that's how it is explained.

At least, I think that's how it is explained.

4431.189 days ago

Apr 12, 2011 - 11:01 PM

This might be a bit vague, but here goes.

Given: (x+3)^2

Find: f(x+h)

Any help?

Given: (x+3)^2

Find: f(x+h)

Any help?

4431.067 days ago

Apr 13, 2011 - 1:56 AM

Sophrosyne said:

Arcsine undoes Sine. The information you gather from a triangle is the Sine of ABC, which is not useful in finding the angle measurement. We can undo Sine though with Arcsine, just like how the sqrt(x^{2}) == x.

At least, I think that's how it is explained.

At least, I think that's how it is explained.

... but how does putting sine under a sign that means "basically" to give the reciprocal... how does that worK?

4431.061 days ago

Apr 13, 2011 - 2:05 AM

Burrito Famine said:

This might be a bit vague, but here goes.

Given: (x+3)^2

Find: f(x+h)

Any help?

Given: (x+3)^2

Find: f(x+h)

Any help?

4431.058 days ago

Apr 13, 2011 - 2:09 AM

I'm only asking why you get the angle that creates the ratio of Sine when you press Sin^{-1} Yes, I understand what ArcSine is, but how does getting a reciprocal reproduce the angle?

4431.054 days ago

Apr 13, 2011 - 2:15 AM

Ah, it's not a reciprocal. The symbol is a bit misleading. The ^{-1} simply means that it's the inverse function, not that it's the multiplicative inverse (reciprocal).

Edit: In other words, sin^{-1} is just a function designed to return the angle that yields whatever sine value is given it.

Edit: In other words, sin

4431.051 days ago

Apr 13, 2011 - 2:19 AM

The ^{-1} doesn't actually mean the reciprocal - it's notational shorthand for arcsin.

In the same way that x^{-1}(x(a)) = a, sin^{-1}(sin(a)) = a.

Also, to the function question:

If you mean that f(x) = (x + 3)^{2}, then f(x+h) = ((x+h)+3)^{2}. Everywhere you see x in the original function, replace it with (x+h).

In the same way that x

Also, to the function question:

If you mean that f(x) = (x + 3)

4431.051 days ago

Apr 13, 2011 - 2:20 AM

Hydrogen777 said:

Ah, it's not a reciprocal. The symbol is a bit misleading. The ^{-1} simply means that it's the inverse function, not that it's the multiplicative inverse (reciprocal).

so what's an inverse function?

4431.051 days ago

Apr 13, 2011 - 2:20 AM

It is a function that "undoes" a corresponding function. The inverse of the function "f(x) = x + 1" for example would be "f^{-1}(x) = x - 1".

4431.049 days ago

Apr 13, 2011 - 2:23 AM

Hydrogen777 said:

Burrito Famine said:

This might be a bit vague, but here goes.

Given: (x+3)^2

Find: f(x+h)

Any help?

Given: (x+3)^2

Find: f(x+h)

Any help?

So am I. But its on my homework and I haven't the slightest clue on what to do. I don't even know what it is asking. Oh well...

4431.033 days ago

Apr 13, 2011 - 2:46 AM

Burrito Famine said:

This might be a bit vague, but here goes.

Given: (x+3)^2

Find: f(x+h)

Any help?

Given: (x+3)^2

Find: f(x+h)

Any help?

I dunno. If f(x)=(x+3)

f(x+h)=[(x+h)+3]

f(x+h)=(x+h+3)

f(x+h)=x

f(x+h)=x

4430.996 days ago

Apr 13, 2011 - 3:39 AM

If this is just a derivative, when f(x)=(x+3)^2, f'(x) = 2(x+3). Easy :)

4430.454 days ago

Apr 13, 2011 - 4:39 PM

Hoooly shit, I'm so fucked if there is no one here to help me, like now!

4430.293 days ago

Apr 13, 2011 - 8:32 PM

I have homework due tomorrow, 3 questions only, but I don't know shit about them!

And here it's 10:35 pm!

And here it's 10:35 pm!

4430.29 days ago

Apr 13, 2011 - 8:35 PM

marymansour said:

If this is just a derivative, when f(x)=(x+3)^2, f'(x) = 2(x+3). Easy :)

Err, wouldn't it be (x+3)*(x+3), which would make the answer x^2+6x+9? The complications arise when h is introduced. But no matter. I've already turned it in.

4430.239 days ago

Apr 13, 2011 - 9:49 PM

Burrito Famine said:

Err, wouldn't it be (x+3)*(x+3), which would make the answer x^2+6x+9? The complications arise when h is introduced. But no matter. I've already turned it in.

No, it wouldn't. The Power Rule easily proves when f(x) = (x+3)^2, f'(x) = 2(x+3).

As for when h is used, as above, f(x + h) is usually used as part of the formula "lim(h-->0) ( ( f(x+h) - f(x) ) / h )", which, if f(x) = (x+3)^2, will equal 2(x+3).

4430.228 days ago

Apr 13, 2011 - 10:04 PM

Ok, thank you very much. I'm going in for tutorials tomorrow. I obviously don't understand it at all.

4430.221 days ago

Apr 13, 2011 - 10:15 PM

ywkbme said:

I have homework due tomorrow, 3 questions only, but I don't know shit about them!

And here it's 10:35 pm!

And here it's 10:35 pm!

Let us help. O_o

4430.088 days ago

Apr 14, 2011 - 1:27 AM

My math teacher says that if you've got a geometry problem, and you can't make sense of it, draw random lines that pass through at least one point (preferably 2). You'll eventually get a line that'll make you click.

Well, if sin(theta) gives the sine of an angle, then Sin^{-1}(theta) should do the inverse... right?

SporeInsanity said:

Not really a problem but just a question.

Explain how Sine^{-1} turns sine into an angle, usually theta :P

Explain how Sine

Well, if sin(theta) gives the sine of an angle, then Sin

4429.907 days ago

Apr 14, 2011 - 5:48 AM

4419.294 days ago

Apr 24, 2011 - 8:30 PM

Forum > Technical Corner > Math Problem Thread

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"html": "Not really a problem but just a question.<br />\n<br />\nExplain how Sine<sup>-1</sup> turns sine into an angle, usually theta :P",
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"html": "Arcsine undoes Sine. The information you gather from a triangle is the Sine of ABC, which is not useful in finding the angle measurement. We can undo Sine though with Arcsine, just like how the sqrt(x<sup>2</sup>) == x.<br />\n<br />\nAt least, I think that's how it is explained.",
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"html": "This might be a bit vague, but here goes.<br />\n<br />\nGiven: (x+3)^2<br />\n<br />\nFind: f(x+h)<br />\n<br />\nAny help?",
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"html": "<div style=\"margin:20px; background-image:url(/images/light.png);\"><div style=\"border:1px solid #888; padding:5px;\"><a href=\"/users/sophrosyne/\">Sophrosyne</a> said:</div><div style=\"border:1px solid #888; padding:20px;\">Arcsine undoes Sine. The information you gather from a triangle is the Sine of ABC, which is not useful in finding the angle measurement. We can undo Sine though with Arcsine, just like how the sqrt(x<sup>2</sup>) == x.<br />\n<br />\nAt least, I think that's how it is explained.</div></div><br />\n<br />\n... but how does putting sine under a sign that means "basically" to give the reciprocal... how does that worK?",
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"html": "<div style=\"margin:20px; background-image:url(/images/light.png);\"><div style=\"border:1px solid #888; padding:5px;\"><a href=\"/users/burritofamine/\">Burrito Famine</a> said:</div><div style=\"border:1px solid #888; padding:20px;\">This might be a bit vague, but here goes.<br />\n<br />\nGiven: (x+3)^2<br />\n<br />\nFind: f(x+h)<br />\n<br />\nAny help?</div></div>Are you not given the original function f? I'm a little lost as to what is being asked.",
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"html": "I'm only asking why you get the angle that creates the ratio of Sine when you press Sin<sup>-1</sup> Yes, I understand what ArcSine is, but how does getting a reciprocal reproduce the angle?",
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"html": "Ah, it's not a reciprocal. The symbol is a bit misleading. The <sup>-1</sup> simply means that it's the inverse function, not that it's the multiplicative inverse (reciprocal).<br />\n<br />\nEdit: In other words, sin<sup>-1</sup> is just a function designed to return the angle that yields whatever sine value is given it.",
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"html": "The <sup>-1</sup> doesn't actually mean the reciprocal - it's notational shorthand for arcsin.<br />\n<br />\nIn the same way that x<sup>-1</sup>(x(a)) = a, sin<sup>-1</sup>(sin(a)) = a.<br />\n<br />\nAlso, to the function question:<br />\nIf you mean that f(x) = (x + 3)<sup>2</sup>, then f(x+h) = ((x+h)+3)<sup>2</sup>. Everywhere you see x in the original function, replace it with (x+h).",
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"html": "<div style=\"margin:20px; background-image:url(/images/light.png);\"><div style=\"border:1px solid #888; padding:5px;\"><a href=\"/users/hydrogen777/\">Hydrogen777</a> said:</div><div style=\"border:1px solid #888; padding:20px;\">Ah, it's not a reciprocal. The symbol is a bit misleading. The <sup>-1</sup> simply means that it's the inverse function, not that it's the multiplicative inverse (reciprocal).</div></div><br />\nso what's an inverse function?",
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"html": "It is a function that "undoes" a corresponding function. The inverse of the function "f(x) = x + 1" for example would be "f<sup>-1</sup>(x) = x - 1".",
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"html": "<div style=\"margin:20px; background-image:url(/images/light.png);\"><div style=\"border:1px solid #888; padding:5px;\"><a href=\"/users/hydrogen777/\">Hydrogen777</a> said:</div><div style=\"border:1px solid #888; padding:20px;\"><div style=\"margin:20px; background-image:url(/images/light.png);\"><div style=\"border:1px solid #888; padding:5px;\"><a href=\"/users/burritofamine/\">Burrito Famine</a> said:</div><div style=\"border:1px solid #888; padding:20px;\">This might be a bit vague, but here goes.<br />\n<br />\nGiven: (x+3)^2<br />\n<br />\nFind: f(x+h)<br />\n<br />\nAny help?</div></div>Are you not given the original function f? I'm a little lost as to what is being asked.</div></div><br />\n<br />\nSo am I. But its on my homework and I haven't the slightest clue on what to do. I don't even know what it is asking. Oh well...",
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"html": "If this is just a derivative, when f(x)=(x+3)^2, f'(x) = 2(x+3). Easy :)",
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"html": "<div style=\"margin:20px; background-image:url(/images/light.png);\"><div style=\"border:1px solid #888; padding:5px;\"><a href=\"/users/burritofamine/\">Burrito Famine</a> said:</div><div style=\"border:1px solid #888; padding:20px;\">Err, wouldn't it be (x+3)*(x+3), which would make the answer x^2+6x+9? The complications arise when h is introduced. But no matter. I've already turned it in.</div></div><br />\n<br />\nNo, it wouldn't. The Power Rule easily proves when f(x) = (x+3)^2, f'(x) = 2(x+3).<br />\n<br />\nAs for when h is used, as above, f(x + h) is usually used as part of the formula "lim(h-->0) ( ( f(x+h) - f(x) ) / h )", which, if f(x) = (x+3)^2, will equal 2(x+3).<br />\n<br />\n<img src=\"/uploads/files/172114.jpg\" alt=\"172114.jpg\" />",
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