if cosθ=2/3 and tanθ<0 find the exact value

sin(θ-5π/3)

sin2θ

cos2θ

cos(θ/2)

can some1 please help me?

Solution:Step1First replace x with (x-Δx) in the given funicton f(x) to obtainf(x+Δx) = 2(x+Δx)+1 = 2x + 2Δx + 1 Step2Rewrite f(x-Δx) as (2x + 2Δx + 1) and f(x) as (2x + 1) into the definition of the derivative limit to obtain: lim [(2x + 2Δx + 1) - (2x+1)] / ΔxΔx-> 0 Step 3Combine like terms on the numerator and the reduced expression becomes lim [2Δx] / ΔxΔx-> 0 Step 4Cancel out Δx from both numerator and denominator and finally you get the derivative Derivative of f (x) = 2x+1 = 2 -Note there was no need to substitute Δx with 0 since it was canceled out.

To find the values, we will use the given information:

1. Given: cosθ = 2/3 and tanθ < 0
From the given values, we can deduce that θ is in the second quadrant, where cosine is positive (2/3) and tangent is negative.

2. Let's find the remaining trigonometric functions:

a. sinθ: In the second quadrant, sinθ is positive. To find the value of sinθ, we can use the Pythagorean identity:
sin²θ + cos²θ = 1
sin²θ + (2/3)² = 1
sin²θ + 4/9 = 1
sin²θ = 1 - 4/9
sin²θ = 9/9 - 4/9
sin²θ = 5/9
sinθ = √(5/9) = √5/3

b. cosθ: Already given as 2/3.

3. Now, let's solve the additional trigonometric expressions:

a. sin(θ - 5π/3): We can use the angle subtraction formula for sine:
sin(θ - 5π/3) = sinθ * cos(5π/3) - cosθ * sin(5π/3)
sin(θ - 5π/3) = (√5/3) * (-1/2) - (2/3) * (√3/2)
sin(θ - 5π/3) = -√5/6 - √3/3

b. sin2θ: We can use the double-angle formula for sine:
sin2θ = 2 * sinθ * cosθ
sin2θ = 2 * (√5/3) * (2/3)
sin2θ = 4√5/9

c. cos2θ: We can use the double-angle formula for cosine:
cos2θ = cos²θ - sin²θ
cos2θ = (2/3)² - (√5/3)²
cos2θ = 4/9 - 5/9
cos2θ = -1/9

d. cos(θ/2): We can use the half-angle formula for cosine:
cos(θ/2) = ±√(1 + cosθ) / 2
Since cosθ = 2/3 is positive, cos(θ/2) is also positive:
cos(θ/2) = √(1 + 2/3) / 2
cos(θ/2) = √(3/3 + 2/3) / 2
cos(θ/2) = √(5/3) / 2

So, the exact values are:
- sin(θ - 5π/3) = -√5/6 - √3/3
- sin2θ = 4√5/9
- cos2θ = -1/9
- cos(θ/2) = √(5/3) / 2