Small-angle approximation: Difference between revisions

Content deleted Content added
Slide-rule approximations: this parenthetical is unnecessary and distracting. slightly re-order last paragraph to flow better
Geometric: there should be a new paragraph after the picture. try to this up a bit (could perhaps still use work)
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=== Geometric ===
[[File:Small angle triangle.svg|400pxframeless|upright=2]]
The red section on the right, {{math|d}}, is the difference between the lengths of the hypotenuse, {{mvar|H}}, and the adjacent side, {{mvar|A}}. As is shown, {{mvar|H}} and {{mvar|A}} are almost the same length, meaning {{math|cos ''θ''}} is close to 1 and {{math|{{sfrac|''θ''<sup>2</sup>|2}}}} helps trim the red away.
<math display="block"> \cos{\theta} \approx 1 - \frac{\theta^2}{2}</math>
 
For a small angle, {{mvar|H}} and {{mvar|A}} are almost the same length, and therefore {{math|cos ''θ''}} is nearly 1. The segment {{mvar|d}} (in red to the right) is the difference between the lengths of the hypotenuse, {{mvar|H}}, and the adjacent side, {{mvar|A}}, and has length <math>\textstyle H - \sqrt{H^2 - O^2}</math>, which for small angles is approximately equal to <math>\tfrac12 \theta^2H</math>. As a second order approximation,
The opposite leg, {{mvar|O}}, is approximately equal to the length of the blue arc, {{mvar|s}}. Gathering facts from geometry, {{math|1=''s'' = ''Aθ''}}, from trigonometry, {{math|1=sin ''θ'' = {{sfrac|''O''|''H''}}}} and {{math|1=tan ''θ'' = {{sfrac|''O''|''A''}}}}, and from the picture, {{math|''O'' ≈ ''s''}} and {{math|''H'' ≈ ''A''}} leads to:
<math display="block"> \cos{\theta} \approx 1 - \frac{\theta^2}{2}.</math>
 
The opposite leg, {{mvar|O}}, is approximately equal to the length of the blue arc, {{mvar|s}}. GatheringThe factsarc from{{mvar|s}} geometry,has length {{math|1=''s'' = ''AθθA''}}, fromand trigonometry,by definition {{math|1=sin ''θ'' = {{sfrac|''O''|''H''}}}} and {{math|1=tan ''θ'' = {{sfrac|''O''|''A''}}}}, and fromfor thea picturesmall angle, {{math|''O'' ≈ ''s''}} and {{math|''H'' ≈ ''A''}}, which leads to:
<math display="block">\sin \theta = \frac{O}{H}\approx\frac{O}{A} = \tan \theta = \frac{O}{A} \approx \frac{s}{A} = \frac{A\theta}{A} = \theta.</math>
 
Or, more concisely,
Simplifying leaves,
<math display="block">\sin \theta \approx \tan \theta \approx \theta.</math>