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January 5 Information
The sine of {0, pi/6, pi/4, pi/3, pi/2} is {root 0 over 2, root 1 over 2, root 2 over 2, root 3 over 2, root 4 over 2}. Thank you. 69.22.242.15 ( talk) 20:24, 5 January 2017 (UTC)
- Because of the Pythagorean theorem? -- Jayron 32 20:35, 5 January 2017 (UTC)
- Substitute that into the formula cos 2A = 1 - 2 sin2A and you get:
- cos {0, π / 3, π / 2, 2π / 3, π} is {2/2, 1/2, 0/2, -1/2, -2/2}
- does that look more familiar or obvious?--
JohnBlackburne
words
deeds
20:39, 5 January 2017 (UTC)
- It does. It looks like the interaction between exponents of 2 and 1. Does it have anything to do with Spiral_of_Theodorus? 69.22.242.15 ( talk) 21:08, 5 January 2017 (UTC)
- I think the OP was wondering if there is any intuition that would lead us to predict such a pattern before using the Pythagoren theorem on each one individually.
Loraof (
talk)
20:54, 5 January 2017 (UTC)
- The answer I give is sort of intuitive; you can draw it as a circle, radius 1, with lines at 0, ±1, ± 1/2. Measure the angles from the top to get {0, π / 3, π / 2, 2π / 3, π}. By trigonometry the cosines of these are the heights of the lines. Finally plug those into
- sin A = √1/2 (1 - cos 2A)
- to get the numbers in the original question.-- JohnBlackburne words deeds 21:11, 5 January 2017 (UTC)
- The answer I give is sort of intuitive; you can draw it as a circle, radius 1, with lines at 0, ±1, ± 1/2. Measure the angles from the top to get {0, π / 3, π / 2, 2π / 3, π}. By trigonometry the cosines of these are the heights of the lines. Finally plug those into
| Mathematics desk | ||
|---|---|---|
| < January 4 | << Dec | January | Feb >> | January 6 > |
| Welcome to the Wikipedia Mathematics Reference Desk Archives |
|---|
| The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
January 5 Information
The sine of {0, pi/6, pi/4, pi/3, pi/2} is {root 0 over 2, root 1 over 2, root 2 over 2, root 3 over 2, root 4 over 2}. Thank you. 69.22.242.15 ( talk) 20:24, 5 January 2017 (UTC)
- Because of the Pythagorean theorem? -- Jayron 32 20:35, 5 January 2017 (UTC)
- Substitute that into the formula cos 2A = 1 - 2 sin2A and you get:
- cos {0, π / 3, π / 2, 2π / 3, π} is {2/2, 1/2, 0/2, -1/2, -2/2}
- does that look more familiar or obvious?--
JohnBlackburne
words
deeds
20:39, 5 January 2017 (UTC)
- It does. It looks like the interaction between exponents of 2 and 1. Does it have anything to do with Spiral_of_Theodorus? 69.22.242.15 ( talk) 21:08, 5 January 2017 (UTC)
- I think the OP was wondering if there is any intuition that would lead us to predict such a pattern before using the Pythagoren theorem on each one individually.
Loraof (
talk)
20:54, 5 January 2017 (UTC)
- The answer I give is sort of intuitive; you can draw it as a circle, radius 1, with lines at 0, ±1, ± 1/2. Measure the angles from the top to get {0, π / 3, π / 2, 2π / 3, π}. By trigonometry the cosines of these are the heights of the lines. Finally plug those into
- sin A = √1/2 (1 - cos 2A)
- to get the numbers in the original question.-- JohnBlackburne words deeds 21:11, 5 January 2017 (UTC)
- The answer I give is sort of intuitive; you can draw it as a circle, radius 1, with lines at 0, ±1, ± 1/2. Measure the angles from the top to get {0, π / 3, π / 2, 2π / 3, π}. By trigonometry the cosines of these are the heights of the lines. Finally plug those into