What is the exact value of sin(2 theta) if cos (theta)=3/5 and (theta) is in quadrent 4?
12/25
24/25
-12/25
-7/25
-24/25
sin2θ = 2sinθcosθ
in QIV, if cosθ = 3/5, sinθ = -4/5
so, sin2θ = 2(-4/5)(3/5) = -24/25
To find the exact value of sin(2θ), we will use the double-angle formula for sine, which states that sin(2θ) = 2sin(θ)cos(θ).
Given that cos(θ) = 3/5, we can determine the value of sin(θ) using the Pythagorean identity sin²(θ) + cos²(θ) = 1.
Since θ is in quadrant 4, which is the bottom right quadrant of the unit circle, we know that sin(θ) is negative. Thus, we have sin(θ) = -√(1 - cos²(θ)).
Substituting the value of cos(θ) = 3/5 into the equation, we get sin(θ) = -√(1 - (3/5)²) = -√(1 - 9/25) = -√(25/25 - 9/25) = -√(16/25) = -4/5.
Now, we can find sin(2θ) using the double-angle formula: sin(2θ) = 2sin(θ)cos(θ).
Substituting in sin(θ) = -4/5 and cos(θ) = 3/5, we get sin(2θ) = 2(-4/5)(3/5) = (-8/5)(3/5) = -24/25.
Therefore, the exact value of sin(2θ) is -24/25.