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# Why is sin of theta appromately equal to theta

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sinΘ ≈Θ

Guest Aug 9, 2014

#3
+91928
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I'd like to take Alan's explanation and expand upon it a little.

NOTE: THIS IS ONLY TRUE AS THE ANGLE TENDS TO 0

$$The circumference of a circle is 2\pi r\\ If the radius is 1 unit then the circumference is 2\pi\\ So if the angle is measured in radians then the arc length will be equal to the angle that it is subtended from. This is for a unit circle.\\\\$$

Melody  Aug 9, 2014
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#1
+4471
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AzizHusain  Aug 9, 2014
#2
+26547
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The following images might help:

So sin θ ≈ θ only when θ is small

Alan  Aug 9, 2014
#3
+91928
+8

I'd like to take Alan's explanation and expand upon it a little.

NOTE: THIS IS ONLY TRUE AS THE ANGLE TENDS TO 0

$$The circumference of a circle is 2\pi r\\ If the radius is 1 unit then the circumference is 2\pi\\ So if the angle is measured in radians then the arc length will be equal to the angle that it is subtended from. This is for a unit circle.\\\\$$

Melody  Aug 9, 2014
#4
+84166
+5

Here's the proof of this using the "Squeeze Theorem".....it's somewhat complicated, but if you have some patience, I think you will appreciate the geometry involved. The proof is the first one on the page.

http://tutorial.math.lamar.edu/Classes/CalcI/ProofTrigDeriv.aspx

Thanks to Paul's Online Math Notes for this one.....(I certainly couldn't have derived it!!!)

BTW....."Paul" is Dr. Paul Dawkins...he's a professor at Lamar University in Texas

CPhill  Aug 9, 2014

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