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February 28, 2015

February 28, 2015

Posted by **Yulieth** on Sunday, November 20, 2011 at 8:24pm.

sin2θ

cos( θ-4π/3)

sin(θ/2)

can some1 explain to me how to do these?

- math -
**Henry**, Tuesday, November 22, 2011 at 8:11pmsinA = -5/13 = Y/r,

X^2 + Y^2 = r^2,

X^2 + (-5)^2 = (13)^2,

X^2 + 25 = 169,

X^2 = 169 - 25 = 144,

X = +- 12.

X = -12 Because it places our resultant

vector in the required quadrant(Q3).

(X,Y) = (-12,-5).

tanA = Y/X = --5/-12 = 0.416666,

A = 22.62 Deg. ,

A = 22.62 + 180 = 202.62(Q3).

180 < 202.62 < 270. Q3.

sin(2A) = sin(405.24) = 0.710.

cos(A-4pi/3) = cos(202.62-240) = 0.7946

sin(A/2) = +- sqrt((1-cosA)/2) =

sqrt((1-cos202.62)/2) = 0.9806

- math -
**Henry**, Wednesday, November 23, 2011 at 3:25pmCorrection:

sinA = -5/13 = Y/r,

A = 157.38 Deg.

sin(2A) = sin(314.76) = -0.710.

cos(A-4pi/3)=cos(157.38-240)=0.1284

sin(A/2) = +- sqrt((1-cosA)/2) =

sqrt((1-cos157.38)/2 = 0.9806.

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