Understanding the Behavior of g(x) = arcsin(sin(x)) in Trigonometry - Explained!

In summary: Therefore the pythagorean identity is proven. In summary, the conversation is discussing a trigonometry review problem involving the function g(x) = arcsin(sin(x)). The person asking for help is unsure of how to explain the appearance of the graph and is looking for assistance using the first and second derivative tests. Another person provides a detailed explanation using derivatives and the pythagorean identity to show that the graph of this function is a straight line.
  • #1
The_Brain
42
3
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
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
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 I am right, can anyone gimme a sec opinion for this gentlemen, please. I am 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
 

1. What is trigonometry?

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles.

2. How is trigonometry used?

Trigonometry is used to solve problems related to triangles, such as finding missing sides or angles, and in real-life applications such as navigation, engineering, and physics.

3. What are the basic trigonometric functions?

The basic trigonometric functions are sine, cosine, and tangent, which represent the ratios of the sides of a right triangle.

4. How do you solve a trigonometry problem?

To solve a trigonometry problem, you need to identify the given information, determine which trigonometric function to use, and apply the appropriate formula to find the unknown value.

5. What is the unit circle and how is it related to trigonometry?

The unit circle is a circle with a radius of 1, centered at the origin on a coordinate plane. It is used in trigonometry to define the values of the trigonometric functions for any angle, making it a useful tool for solving trigonometry problems.

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