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Given that sinθ= -4/5 and π<θ<3π/2, find the exact values of the following

sin2θ

cos2θ

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Answer:

We can use the double angle formulas to find sin(2θ) and cos(2θ):

sin(2θ) = 2sin(θ)cos(θ)

cos(2θ) = cos^2(θ) - sin^2(θ)

Since sin(θ) = -4/5 and θ is in the third quadrant (π < θ < 3π/2), we can use the Pythagorean identity to find cos(θ):

sin^2(θ) + cos^2(θ) = 1

cos^2(θ) = 1 - sin^2(θ)

cos(θ) = -√(1 - sin^2(θ))

cos(θ) = -√(1 - (-4/5)^2) = -√(1 - 16/25) = -√(9/25) = -3/5

Now we can substitute these values into the double angle formulas:

sin(2θ) = 2sin(θ)cos(θ) = 2(-4/5)(-3/5) = 24/25

cos(2θ) = cos^2(θ) - sin^2(θ) = (-3/5)^2 - (-4/5)^2 = 9/25 - 16/25 = -7/25

Therefore, the exact values of sin(2θ) and cos(2θ) are sin(2θ) = 24/25 and cos(2θ) = -7/25, respectively.

Finally, we can find the value of sin^2(2θ)cos^2(2θ):

sin^2(2θ)cos^2(2θ) = (sin(2θ))^2(cos(2θ))^2

sin^2(2θ)cos^2(2θ) = (24/25)^2(-7/25)^2

sin^2(2θ)cos^2(2θ) = 24^2/25^2 * 7^2/25^2

sin^2(2θ)cos^2(2θ) = (24*7)^2/25^4

Therefore, the exact value of sin^2(2θ)cos^2(2θ) is (24*7)^2/25^4.

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