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What is the distance between the points (\cos 37^{\circ}, \sin 37^{\circ}) and (\cos 127^{\circ}, \sin 127^{\circ})?

 Jun 20, 2025
 #1
avatar+22 
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For points P₁ = (cos 37°, sin 37°) and P₂ = (cos 127°, sin 127°), the distance is:

√((cos 127° - cos 37°)² + (sin 127° - sin 37°)²).

The angle difference: 127° - 37° = 90°. Simplify:

(cos 127° - cos 37°)² + (sin 127° - sin 37°)².

Expand:

= cos² 127° - 2 cos 127° cos 37° + cos² 37° + sin² 127° - 2 sin 127° sin 37° + sin² 37°.

Since cos² θ + sin² θ = 1:

= (1 + 1) - 2 (cos 127° cos 37° + sin 127° sin 37°).

Use the cosine angle sum identity:

cos 127° cos 37° + sin 127° sin 37° = cos (127° - 37°) = cos 90° = 0.

Thus:

= 2 - 2 · 0 = 2.

Distance:

√2.

 Jun 20, 2025
edited by mercurym999  Jun 20, 2025
 #2
avatar+130503 
+1

THX, mercurym999 !!!

 

Another method using the Law of Cosines  (and assuming a radius of 1 )

 

d^2  =  1^2  + 1^2  - 2 ( 1 * 1)  cos (127° -37°)

 

d^2  = 2  - 2  cos (90°)

 

d^2  = 2 - 2 (0)

 

d^2  = 2

 

d =  sqrt 2

 

cool cool cool

 Jun 20, 2025
edited by CPhill  Jun 20, 2025

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