given that cos(theta)= (3/5) and sin (theta)<0,

find sin(2 theta)
cos (2 theta)
tan (2theta)

To find sin(2θ), cos(2θ), and tan(2θ) when you are given cos(θ) and sin(θ), you can use trigonometric identities. In this case, we are given that cos(θ) = 3/5 and sin(θ) < 0.

First, let's find sin(θ). We know that sin²(θ) + cos²(θ) = 1, so substituting the given value of cos(θ), we have sin²(θ) + (3/5)² = 1. Solving this equation gives us sin²(θ) = 1 - (9/25). Simplifying further, sin²(θ) = 16/25. Since sin(θ) < 0, we take the negative square root to get sin(θ) = -4/5.

Next, let's find cos(θ). We already know that cos(θ) = 3/5.

Now, we can use the double angle formulas to find sin(2θ), cos(2θ), and tan(2θ).

1. Sin(2θ) = 2sin(θ)cos(θ)
By substituting the values of sin(θ) and cos(θ) that we found, we have sin(2θ) = 2 * (-4/5) * (3/5) = -24/25.

2. Cos(2θ) = cos²(θ) - sin²(θ)
By substituting the values of sin(θ) and cos(θ), we have cos(2θ) = (3/5)² - (-4/5)² = 9/25 - 16/25 = -7/25.

3. Tan(2θ) = sin(2θ) / cos(2θ)
By substituting the values of sin(2θ) and cos(2θ), we have tan(2θ) = (-24/25) / (-7/25) = 24/7.

Therefore, sin(2θ) = -24/25, cos(2θ) = -7/25, and tan(2θ) = 24/7.