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March 2, 2015

March 2, 2015

Posted by **Briana** on Friday, April 13, 2012 at 3:17pm.

- Trig -
**Damon**, Friday, April 13, 2012 at 3:30pmsin (30/2) = +/- sqrt [(1-cos 30)/2]

+ in first quadrant

= sqrt [ (1 - {sqrt 3}/2)/2) ]

= sqrt [ (2 - sqrt 3)/4 ]

= (1/2) sqrt (2 - sqrt 3)

- Trig -
**Reiny**, Friday, April 13, 2012 at 3:35pmuse

cos 2A = 1 - 2sin^2 A

cos 30° = 1 - 2sin^2 15°

√3/2 = 1- 2sin^2 15°

2sin^2 15° - 1 - √3/2 = (2-√3)/2

sin^2 15° = (2-√3)/4

sin 15° = √(2-√3) /2

it would have been easier to do

sin15°

= sin(45°-30°)

= sin45cos30 - cos45sin30

= (√2/2)(√3/2) -(√2/2)(1/2)

= (√6 - √2)/4 which yields the same result as above

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