How is the area under the arch of y=sinx related to the unit circle?

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SUMMARY

The discussion centers on the relationship between the area under the sine curve \(y = \sin x\) and the unit circle, particularly in the context of Year 12 mathematics. User Travis explores calculating the area using triangles and trapeziums, arriving at an area of approximately 1.896. The conversation highlights the importance of visual aids, such as auxiliary lines, in understanding geometric relationships. Additionally, Travis seeks clarification on the derivation of the formula \( \frac{\pi}{4}(1+\sqrt{2})^3 \) and its connection to the area under the sine curve.

PREREQUISITES
  • Understanding of basic trigonometry, specifically sine functions
  • Familiarity with geometric shapes, including triangles and trapeziums
  • Knowledge of area calculation formulas for triangles and trapeziums
  • Concept of the unit circle and its relationship to trigonometric functions
NEXT STEPS
  • Study the derivation of the formula \( \frac{\pi}{4}(1+\sqrt{2})^3 \) in relation to sine curves
  • Learn about numerical integration techniques for calculating areas under curves
  • Explore the properties of the unit circle and its implications for trigonometric functions
  • Investigate the relationship between sine functions and their graphical representations
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Students preparing for Year 12 mathematics, educators teaching trigonometry and geometry, and anyone interested in the mathematical properties of sine functions and their applications.

Trav44
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Hi guys I'm new to the forum but with year 12 maths coming up next year I assume I'll be coming here a lot.

This question is part of my end of year investigation.

I have scanned the questions and attached them as jpegs.

I was thinking something along the lines of using the triangle to determine the area? Or if the fact that this curve is related to half of a unit circle had something to do with it?

Any help would be much appreciated :)

Thank you, Travis
 

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You've got to show an attempt at solving the problem in order to get some help.
 
Hi Trav, welcome to the forums. I do hope you'll be coming here - and learning - a lot.
As SteamKing said, please show us your attempt or problem. The questions are leading you in a specific direction. As you've noticed, there is a pretty big hint in the picture in the form of the auxiliary lines that have been drawn, creating for example a triangle. If you show us how far you got with that, we may be able to help you complete it.
 
Sorry I was unaware that an attempt must be shown, I will be sure to include them in every question from here on out :)
My attempts for each question as follows are:
1. The area of the outer rectangle is pi as pi x 1 = pi
The area for the inner triangle is 1/2bh so 1/2 x pi x 1= 1/2pi or pi/2
this shows that the area under the sine arch must fit the range of pi./2 < x < pi

2.
Triangle area= 1/2 x b x h
1/2 x pi/4 x sin(pi/4)= 0.2776

Trapezium area = (a + b)/2 x h
(sin (pi/4) + sin(pi/2))/2 x pi/4=0.6704

total area = 0.2776x2 + 0.6704x2= 1.896

which is the correct area

I get that sin of (pi/4 or pi/2) is the height at those given points but could someone please give me an explanation as to why this is?

I have found the area for both triangles and trapeziums and multiplied them by 2 to give me the overall area and this is right how ever I cannot make the connection between this and the equation pi/4(1+(root 2))3. I can use the formula to add up all the areas and I get an answer close to 2 which seems logical as the area under a sine curve is meant to equal 2.
I do not understand how this formula has been derived though?
And for pi to be split into increments of 0.1 I thought the start of the equation should be 1/31.4 as this is pi/0.1?
1/20 would give increments close to 0.15 wouldn't it?
 

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