SOLUTION: If sinθ= 4/5 and θ terminates in Quadrant II, find the exact value of sin2θ. F. -24/25 G. 8/5 (I chose this one and it was wrong) H.-12/25 J. 24/25 Sorry I foun

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Question 825715: If sinθ= 4/5 and θ terminates in Quadrant II, find the exact value of sin2θ.
F. -24/25
G. 8/5 (I chose this one and it was wrong)
H.-12/25
J. 24/25
Sorry I found the typo I submitted in one of my previous questions. So sorry about that.
Found 2 solutions by jim_thompson5910, MathTherapy:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
I'm going to use x in place of θ

sin(x) = 4/5

sin^2(x) = 16/25 ... square both sides

-------------------------------------------------------

sin^2(x) + cos^2(x) = 1

16/25 + cos^2(x) = 1

cos^2(x) = 1 - 16/25

cos^2(x) = 25/25 - 16/25

cos^2(x) = (25 - 16)/25

cos^2(x) = 9/25

cos(x) = -sqrt(9/25) ... we are in quadrant II, so cos(x) < 0

cos(x) = -3/5

--------------------------------------------------------

sin(2x) = 2*sin(x)*cos(x)

sin(2x) = 2*(4/5)*(-3/5)

sin(2x) = (2/1)*(4/5)*(-3/5)

sin(2x) = (2*4*(-3))/(1*5*5)

sin(2x) = -24/25

Final Answer: sin(2x) = -24/25
Answer by MathTherapy(10553)   (Show Source): You can put this solution on YOUR website!
If sinθ= 4/5 and θ terminates in Quadrant II, find the exact value of sin2θ.
F. -24/25
G. 8/5 (I chose this one and it was wrong)
H.-12/25
J. 24/25
Sorry I found the typo I submitted in one of my previous questions. So sorry about that. 


sin θ =  = 

Since the opposite side (O) is 4 and the hypotenuse (H)is 5, the adjacent side (A) MUST = 3,

as this represents a 3-4-5 special triangle



Since θ terminates in the 2nd quadrant, where cos is < 0 (negative), we get:

cos θ =  



sin 2 θ = 2 sin θ cos θ (double-angle identity)

Therefore, sin 2 θ = ________ (CHOICE H) 

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