SOLUTION: Hi there, could someone please help me with this Advanced Calculus, Mathematical Modelling question as I have no idea how to solve it. Going through examples havent helped me at al

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Question 203105: Hi there, could someone please help me with this Advanced Calculus, Mathematical Modelling question as I have no idea how to solve it. Going through examples havent helped me at all.
Question: Find the regions where sinx ≤ x.(3 marks)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Algebraically it is difficult to solve this inequality. But with Geometry and logic we can figure this out. Below is a drawing of a unit circle (a circle with radius of 1).

Let's look at the following cases:
  • x+%3E=+1: sin%28x%29+%3C=+x since sin%28x%29+%3C=+1 for all x.
  • x+%3C+-1: sin%28x%29+%3E+x since sin(x) > -1 for all x.
  • x+=+0: sin%28x%29+=+x since sin%280%29+=+0
  • 0+%3C+x+%3C+1: With x (the radian measure of an angle) between 0 and 1 we must be in the first quadrant. Also, since the vertical segment is the "opposite" side and since the hypotenuse (radius) is 1, the length of the vertical segment is sin(x). From the drawing above we can see that the length of vertical segment must be less than the length of the arc, which is x. So sin%28x%29+%3C+x.
  • -1+%3C=+x+%3C+0: With similar logic as in the previous case, we can determine that the arc length is still longer than the vertical segment. However "longer" negative numbers are less than "shorter" negative numbers. For example: -2%2F3+%3C+-1%2F2. So sin%28x%29+%3E+x.

In summary, sin%28x%29+%3C=+x in the following cases:
  • x+=+0.
  • 0+%3C+x+%3C+1.
  • x+%3E=+1.

In short sin%28x%29+%3C=+x for all x+%3E=+0.