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?
Of course, I can help you with these trigonometry questions!
First, let's find the exact value of sin(θ-5π/3) using the given information.
To find sin(θ-5π/3), we can use the identity sin(A-B) = sinAcosB - cosAsinB.
We know that cosθ = 2/3. Using the Pythagorean identity, we can find sinθ as follows:
sinθ = √(1 - cos²θ)
= √(1 - (2/3)²)
= √(1 - 4/9)
= √(5/9)
= √5/3
Now, let's plug in the values into the identity sin(θ-5π/3):
sin(θ-5π/3) = sinθ*cos(5π/3) - cosθ*sin(5π/3)
We know that cos(5π/3) = -1/2 and sin(5π/3) = √3/2:
sin(θ-5π/3) = (√5/3)*(-1/2) - (2/3)*(√3/2)
= -(√5/6) - (√6/6)
= -((√5 + √6)/6)
So, the exact value of sin(θ-5π/3) is -((√5 + √6)/6).
Moving on to the next question, let's find the exact value of sin2θ.
The identity sin2θ = 2sinθcosθ relates the double-angle of sinθ to its single-angle counterpart.
We already know that sinθ = √5/3. Plugging this value into the identity:
sin2θ = 2*(√5/3)*(2/3)
= (2√5)/9
So, the exact value of sin2θ is (2√5)/9.
Next, let's find the exact value of cos2θ.
The identity cos2θ = cos²θ - sin²θ relates the double-angle of cosθ to its single-angle counterpart.
Using the values we found earlier, cosθ = 2/3 and sinθ = √5/3:
cos2θ = (2/3)² - (√5/3)²
= 4/9 - 5/9
= -1/9
Thus, the exact value of cos2θ is -1/9.
Lastly, let's find the exact value of cos(θ/2).
The identity cos(θ/2) = ±√((1 + cosθ)/2) relates the half-angle of cosθ to its full-angle counterpart.
Using the given value cosθ = 2/3:
cos(θ/2) = ±√((1 + (2/3))/2)
= ±√(5/6)
Since tanθ < 0 (negative), we can conclude that cosθ is negative. Therefore, we take the negative sign:
cos(θ/2) = -√(5/6)
Hence, the exact value of cos(θ/2) is -√(5/6).
I hope this helps! Let me know if you have any further questions.