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EXPLAIN trig. problem

  • Thread starter The_Brain
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  • #1
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Hey, I need a little help on this trigonometry review problem in my calculus book that I can't seem to figure out. It goes something like this...

- Graph the function g(x) = arcsin(sin(x)). How do you explain the appearance of this graph?


This is not a "explain what the graph looks like" question, it is a question asking WHY the graph is like it is and I have no idea. Any help is appreciated, thanks.
 

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  • #2
Tom Mattson
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Originally posted by The_Brain
- Graph the function g(x) = arcsin(sin(x)). How do you explain the appearance of this graph?


This is not a "explain what the graph looks like" question, it is a question asking WHY the graph is like it is (snip)
Sorry, but I don't see the problem.

"How do you explain the appearance of the graph?"

and

"Why does the graph look like it does?"

are, to my mind, the same question. It seems clear that they are looking for you to use the first- and second derivative tests.
 
  • #3
Dx
Originally posted by The_Brain
Hey, I need a little help on this trigonometry review problem in my calculus book that I can't seem to figure out. It goes something like this...

- Graph the function g(x) = arcsin(sin(x)). How do you explain the appearance of this graph?


This is not a "explain what the graph looks like" question, it is a question asking WHY the graph is like it is and I have no idea. Any help is appreciated, thanks.
Do you have a graphing calculator? I believe that arcsin is the inverse of sin so this should be a striaght line. also memorize that sin(90 degrees) = 1 and cos(90) = 0 there are many others like sin(45)=.707, etc... but this will help you out later in other chapters, ok.
I hope Im right, can anyone gimme a sec opinion for this gentlemen, plz. Im about 85% right, i think!
Dx :wink:
 
  • #4
If y = sin(x),

Then y = arcsin(x) is the same graph as x = sin(y)

Arcsin(x) is the inverse of sin(x) and therefore:

Sin(arcsin(x)) and Arcsin(sin(x)) both equal x...

Hence the answer that you get...

Now using derivatives to do the same job:

[itex]\frac{d}{dx}[/itex] [Arcsin(sin(x))] = 1

To solve this problem you need to have memorized (or deducted) the following formula

[itex]\frac{d}{du}[/itex] Arcsin(u) = [itex]\frac{du}{\sqrt{1-u^{2}}}[/itex]

So here if you make the subsitution u = sin(x), du = cos(x), u2 = sin2(x)

Then the problem [itex]\frac{d}{dx}[/itex] [Arcsin(sin(x))]

Now is equal to [itex]\frac{d}{du}[/itex] Arcsin(u)

Which (using the formula) equals: [itex]\frac{du}{\sqrt{1-u^{2}}}[/itex]

So if we make back our subsitutions we get:

[itex]\frac{cos(x)}{\sqrt{1-sin^{2}(x)}}[/itex]

Now we know that 1 - sin2(x) = cos2(x) through the pythagorean identity

So [itex]\frac{cos(x)}{\sqrt{1-sin^{2}(x)}}[/itex] = [itex]\frac{cos(x)}{\sqrt{cos^{2}(x)}}[/itex] = [itex]\frac{cos(x)}{\pm cos(x)}[/itex] = [itex]\pm[/itex]1

The only functions whose derivative are [itex]\pm[/itex]1 are y = [itex]\pm[/itex]x + C where c is any arbitrary constant...

Proof of that is as follows:

[itex]\int[/itex] [itex]\pm[/itex]1 dx = [itex]\pm[/itex]x + C

So all in all if you solved the problem using derivatives it will be as follows:

y = arcsin(sin(x))
u = sin(x) du = cos(x) u2 = sin2(x)
y = arcsin(u)
[itex]\frac{dy}{du}[/itex] = [itex]\frac{du}{\sqrt{1-u^{2}}}[/itex]
Now using substitution:
[itex]\frac{dy}{dx}[/itex] = [itex]\frac{cos(x)}{\sqrt{1-sin^{2}(x)}}[/itex]
Pythagorean Identity:
[itex]\frac{cos(x)}{\sqrt{1-sin^{2}(x)}}[/itex] = [itex]\frac{cos(x)}{\sqrt{cos^{2}(x)}}[/itex] = [itex]\frac{cos(x)}{\pm cos(x)}[/itex] = 1
So [itex]\frac{dy}{dx}[/itex] = [itex]\pm[/itex]1
Meaning y = [itex]\pm[/itex]x + c
Notice that arcsin(sin(0)) = 1 and acrsin(sin(1)) = 1 therefore,
y = x
 
Last edited:
  • #5
Btw for the follow up proof:

[itex]\frac{d}{dx}[/itex] arcsin(x) = [itex]\frac{dx}{\sqrt{1-x^{2}}}[/itex]

because of the following:

y = arcsin(x) is the same thing as saying:

x = sin(y)

now [itex]\frac{d}{dx}[/itex] [ x = sin(y) ]

produces the implicit equation: 1 (dx) = cos(y) [itex]\frac{dy}{dx}[/itex]

Which when solved for [itex]\frac{dy}{dx}[/itex] produces: [itex]\frac{dx}{cos(y)}[/itex] = [itex]\frac{dy}{dx}[/itex]

Realize that once again due to the pythagorean identity, cos2(y) = 1 - sin2(y)

so cos(y) = [itex]\sqrt{1 - sin^{2}(y)}[/itex] which substitutes back into the equation

[itex]\frac{dx}{\sqrt{1 - sin^{2}(y)}}[/itex]


Remember our original equation: x = sin(y) so we now substitute x for sin(y) to finally get:

[itex]\frac{dy}{dx}[/itex] = [itex]\frac{dx}{\sqrt{1 - x^{2}}}[/itex]

-----------------------------------------------------------------------------------------

Now both proofs i produced are built on the pythagorean identity which is as follows:

sin2(x) + cos2(x) = 1

Keep in mind the three sides of a right triangle are as follows: legs a and b and hypotenuse C

sin(x) = [itex]\frac{b}{c}[/itex]

cos(x) = [itex]\frac{a}{c}[/itex]

So here is the pythagorean theorem c2 = a2 + b2

sin2(x) + cos2(x) = [itex]\frac{b^{2}}{c^{2}}[/itex] + [itex]\frac{a^{2}}{b^{2}}[/itex] = [itex]\frac{a^{2} + b^{2}}{c^{2}}[/itex]

Now recall that c2 = a2 + b2 so

sin2(x) + cos2(x) = [itex]\frac{c^{2}}{c^{2}}[/itex] = 1

so: sin2(x) + cos2(x) = 1
 

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