What is the exact value of sin(2 theta) if cos theta= 3/5 and theta is in quadrant IV?

To find the exact value of sin(2θ), we can use the double-angle identity formula for sine: sin(2θ) = 2sin(θ)cos(θ).

Given that cos(θ) = 3/5, we need to determine the value of sin(θ) to compute sin(2θ). Since θ is in the fourth quadrant, where cosine is positive and sine is negative, we can use the Pythagorean identity to find sin(θ).

First, let's find the value of sin(θ):
sin²(θ) + cos²(θ) = 1
sin²(θ) + (3/5)² = 1
sin²(θ) + 9/25 = 1
sin²(θ) = 1 - 9/25
sin²(θ) = 25/25 - 9/25
sin²(θ) = 16/25

Since sin(θ) is negative in the fourth quadrant, we take the negative square root:
sin(θ) = -√(16/25)
sin(θ) = -4/5

Now, we can substitute the values of sin(θ) and cos(θ) into the double-angle identity formula for sine:
sin(2θ) = 2sin(θ)cos(θ)
sin(2θ) = 2(-4/5)(3/5)
sin(2θ) = -24/25

Therefore, the exact value of sin(2θ) is -24/25.

in QIV,

cosθ = 3/5
sinθ = -4/5

and you know that

sin2θ = 2sinθ cosθ

So, plug and chug