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July 31, 2014

July 31, 2014

Posted by **Sierra** on Wednesday, November 7, 2012 at 10:51am.

- trig -
**Steve**, Wednesday, November 7, 2012 at 11:08amcos2θ = 1 - 2sin^2 θ

since 2θ = 5pi/4, cos 2θ = -1/√2

-1/√2 = 1 - 2sin^2 5π/8

2sin^2 5π/8 = 1 + 1/√2

sin 5π/8 = √(1 + 1/√2)/2

= √(2+√2) / 2

- trig -
**Damon**, Wednesday, November 7, 2012 at 11:17amYou are talking about 112.5 degrees

that is 90 + 22.5

22.5 is half of 45

so draw this in the third quadrant

I know functions of 45 degrees

sin 45 = +1/sqrt 2

cos 45 = +1/sqrt 2

so what are the functions of 22.5 degrees?

sin(45/2) = +/- sqrt[(1-cos 45)/2]

cos (45/2) = +/-sqrt[(1+cos 45)/2]

in this case we want the negative cos for the sin (look at sketch) because in quadrant 3

so sin 112.5 = -sqrt [ (1+1/sqrt 2)/2]

= -sqrt [ (1 + sqrt 2)/2 sqrt 2

=- sqrt [ (2+sqrt2)/4 ]

= - .923

check sin 112.5 on calculator = -.923

- trig - note -
**Steve**, Wednesday, November 7, 2012 at 11:22amsin is positive in QII

- trig -
**Damon**, Wednesday, November 7, 2012 at 2:52pmwhoops, sorry

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