If sinθ=1/5 and θ is in Quadrant I, find the exact value of sin2θ.
To find the exact value of sin2θ, we need to use the double angle formula for sine. The double angle formula states that sin2θ = 2sinθcosθ.
First, we need to find the value of cosθ. Since θ is in Quadrant I and sinθ = 1/5, we can use the Pythagorean identity to find cosθ. The Pythagorean identity states that sin^2θ + cos^2θ = 1.
Given that sinθ = 1/5, we can substitute that into the identity:
(1/5)^2 + cos^2θ = 1
1/25 + cos^2θ = 1
cos^2θ = 1 - 1/25
cos^2θ = 24/25
Taking the square root of both sides, we get:
cosθ = √(24/25)
Since θ is in Quadrant I, cosθ is positive, so we take the positive square root:
cosθ = √24/√25
cosθ = √24/5
Now that we have sinθ and cosθ, we can use the double angle formula to find sin2θ:
sin2θ = 2sinθcosθ
sin2θ = 2 * (1/5) * (√24/5)
sin2θ = 2√24/25
To find the exact value of sin2θ, we can use the double-angle formula for sine, which states:
sin2θ = 2sinθcosθ
Given that sinθ = 1/5, we need to find the value of cosθ. Since θ is in Quadrant I, which is the positive x-axis, we can use the Pythagorean identity to find cosθ:
cosθ = √(1 - sin^2θ)
Plugging in the given value of sinθ = 1/5:
cosθ = √(1 - (1/5)^2)
= √(1 - 1/25)
= √(24/25)
= √24 / 5
Now, we can substitute the values of sinθ and cosθ into the formula for sin2θ:
sin2θ = 2sinθcosθ
= 2(1/5)(√24 / 5)
= (2/5)(√24 / 5)
= 2√24 / 25
Therefore, the exact value of sin2θ when sinθ = 1/5 and θ is in Quadrant I is 2√24 / 25.