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December 18, 2014

December 18, 2014

Posted by **Yuli** on Monday, November 21, 2011 at 4:44am.

sin2θ

cos( θ-4π/3)

sin(θ/2)

can some1 please help me with this?

- mathematics -
**Steve**, Monday, November 21, 2011 at 5:13amLet's go through this again.

θ is in the 3rd quadrant, so x and y are both negative, and the hypotenuse h = 13

sinθ = -5/13

cosθ = -12/13

Use these values in your half-angle and double-angle and sum/difference formulas.

sin 2θ = 2 sinθ cosθ = 2(-5/13)(-12/13) = 120/169

cos(θ-4π/3) = cosθ cos4π/3 + sinθ sin4π/3

= (-12/13)(-1/2) + (-5/13)(-√3/2)

= 12/26 + 5√3/26

sin θ/2 = √((1-cosθ)/2)

= √((1 + 12/13)/2)

= √(25/26)

Use this same method for all your other similar problems.

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