Expert: Scotto - 4/30/2009

Question
anything you can answer REALLY DOES HELP!

So first part is:

1. to this trig identity, sin(x+h)=sin(x)cos(h)+cos(x)sin(h)..
by using factoring and limit laws how do I simplify this further so that there is no x's in the limits??

2. What is the area of a sector with angle Pi/2 and radius r?

3. Find a formula for the area of a sector with angle (theta) and radius r.
Area of sector= (area of circle)(fraction of circle)=

4. from this picture,(it looks like a triangle with a rounded part at one side as taken from a circle)with theta as the angle and r on both sides. theta is facing the curved circle edge. The question is:

Area of small triangle < area of circle < Area of large triangle

Substitue formulas for the areas into this inequality.

5. After I find answer to 4, it says I need to multiply by 2 to simplify the inequality.

6. says I need to substitute i think t=(cos(theta),sin(theta)) in the inequality to part 5 above.

7. it then says if a<b, what's the relationship between 1/a and 1/b, I said 1/a > 1/b.. is that suppose to be right?

8. use part 6 above to invert the inequality in part f.

9. multiply by sin(theta). you should now have sin(theta)/(theta) between two other quantities.

10. Use the squeeze theorem to find lim    sin(theta)
                               (theta)->0 ---------
                                           (theta)


theres more but I don't want to overload my questions cuz I know its too much to ask but anything you can answer would help me so much. Its a step by step thing i'm suppose to do and if i can't get one part I can't get another part later on that refers to it. So anything really does help.

thank you!  

Answer
1. to this trig identity, sin(x+h)=sin(x)cos(h)+cos(x)sin(h)..
by using factoring and limit laws how do I simplify this further so that there is no x's in the limits??

No.  When h goes to zero, note that cos(h)->1 and sin(h)->0.


2. What is the area of a sector with angle Pi/2 and radius r?
The angles is 1/4th of 2π, so the are is 1/4 of the area of a circle with radius r.


3. Find a formula for the area of a sector with angle (theta) and radius r.
Area of sector= (area of circle)(fraction of circle)=

The area of a circle is πr².
The angle is a fraction of the total.
The total angle is 2π radians.
A = Θπr²/(2π) = Θr²/2.


4. from this picture,(it looks like a triangle with a rounded part at one side as taken from a circle)with theta as the angle and r on both sides. theta is facing the curved circle edge. The question is:

Area of small triangle < area of circle < Area of large triangle

Substitue formulas for the areas into this inequality.

Put one point at the center of where the cirle would be.
Put one point at the end of the triangle if all drawn.
Put one point where the radius meets the triangle.
This is a right triangle, so find the area.  

One side is r, the raidus.
The other sides is given by 2Θ where tanΘ = r/x,
where x is the side that is on the edge of the triangle.

Using this, we know r and 2Θ, so compue Θ, find the tan(), and get x.
That are of the triangle is found by 2Θ.  We know that Θ is half of 2Θ, so the angle at the center of the circle is Φ = π/2 - Θ.
The angle is duplicated on the other side,
so we have 2Φ out of a circle.

Take πr² as the circle area, multiply by Φ/π for that section.
Subtract that from the area of both triangles to get the section outside.


5. After I find answer to 4, it says I need to multiply by 2 to simplify the inequality.

I'm not sure what is meant or if I answered the last one right.
From drawing the picture, the total area can be seen.

6. says I need to substitute i think t=(cos(theta),sin(theta)) in the inequality to part 5 above.

I'm not sure what this is for either.


7. it then says if a<b, what's the relationship between 1/a and 1/b, I said 1/a > 1/b.. is that suppose to be right?

If a and b are both positive it is.
 
You see, dealing with ineqaulities,
when you multiply by a negative, it reverses them.


8. use part 6 above to invert the inequality in part f.

f?  Where is f?


9. multiply by sin(theta). you should now have sin(theta)/(theta) between two other quantities.


10. Use the squeeze theorem to find lim    sin(theta)
                              (theta)->0 ---------
                                          (theta)

Forget all of the above stuff.
The Taylor's theorm for sinΘ is Θ - Θ^3/3! + Θ^5/5! ...
When you divide by Θ, all the terms except the first still have Θ.
As Θ goes to zero, only the first term is left.

Therefore the answer is 1.