If sinθ = 2 over 3 and tanθ < 0, what is the value of cosθ?

Solution: 

Sinθ = 2/3 (Given)

The angle ∠BOA' = 180° - θ° or π - θ (in radians)

Now sin(180° - θ°) = sin(θ)

Sinθ = A'B/OB (by definition i.e. Sinθ = perpendicular/hypotenuse

Sinθ = 2/3

The above problem statement is represented by the figure below:

A'B = 2 (perpendicular)

OB = 3 (Hypotenuse)

Since OA'B is right angled triangle, using the pythagorean theorem we have:

OB² = OA'² + A'B²

OB² = OA'² + A'B²

(3)² = OA'² + (2)²

Hence,

(3)² - (2)² = OA'²

OA'² = 9 - 4 = 5

OA = 5

The value Cosθ = Base is hypotenuse = -√5/3 (because tanθ < 0)

Aliter:

Given, sin θ = 2 / 3. We will find the value of cos θ.

Using the trigonometric identity: sin² θ + cos² θ = 1

Substituting the value of sin θ = 2 / 3, we get:

(2 / 3)² + cos² θ = 1

cos² θ = 1 - (2 / 3)²

cos² θ = 1 - 4 / 9

cos² θ = 5 / 9

cos θ = ±√5/3

cos θ = -√5/3

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If sinθ = 2 over 3 and tanθ < 0, what is the value of cosθ?

Summary:

If sinθ = 2 over 3 and tanθ < 0, then value of Cosθ = -√5/3. In the given problem sinθ is +ve (2/3) and tanθ is less than zero (-ve).