SOLUTION: If secθ= 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

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Question 825698: If secθ= 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
Found 2 solutions by jim_thompson5910, Seutip:
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 (it's easier to type)

sec(x) = 4/5

1/cos(x) = 4/5

cos(x) = 5/4 ... take the reciprocal of both sides

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

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

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

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

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

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

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

sin^2(x) = -9/16

sin(x) = sqrt(-9/16)

As you can see in the last step above, we have sqrt(-9/16), but we cannot take the square root of a negative number and get a real number as a result. So this leads me to think that there's a typo somewhere. Please double check. Thanks.
Answer by Seutip(231)   (Show Source): You can put this solution on YOUR website!
If secθ= 4/5 and θ terminates in Quadrant II, find the exact value of sin2θ.
cos^2θ + sin^2θ = 1 because of Pythagorean identities
Therefore sin^2θ = 1- cos^2θ
cosθ = 1/secθ
So,
sin^2θ = 1- cos^2θ
sin^2θ = 1- (1/secθ)^2
sin^2θ = 1- {1/ (4/5)}^2
sin^2θ = 1- (5/4)^2
sin^2θ = 1- (25/16)
sin^2θ = 1- (25/16)
sin^2θ = 16/16 - (25/16)
sin^2θ = 16-25 /16
sin^2θ = -9/16
Tht's the answer :)
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