Given that tan¤=3/4 and ¤ is in the second quadrant find sin2¤ (¤=theta)
if tan is 3/4, then the right triangle is 3,4,5 so sin is 3/5, second quadrant.
That was a good atempt bob but the options i have here is -2/35, 30.78, -24/35, AND 18
To find sin2θ, we need to use the double-angle identity formula for sine:
sin2θ = 2sinθcosθ
Given that tanθ = 3/4, we can solve for sinθ and cosθ using the trigonometric identity:
tanθ = sinθ/cosθ
Since tanθ is positive in the second quadrant, we know that sinθ and cosθ must have opposite signs. We can assume that sinθ is positive and cosθ is negative.
From the given equation tanθ = 3/4, we can conclude that sinθ = 3 and cosθ = -4. To find these values, we can use the Pythagorean identity:
sin^2θ + cos^2θ = 1
(3^2) + (-4^2) = 1
9 + 16 = 1
25 = 1
This equation is not true, which means that the values we assumed for sinθ and cosθ are incorrect.
Given that tanθ = 3/4 and θ is in the second quadrant, we need to use the Pythagorean identity to find the values of sinθ and cosθ.
Recall that tanθ = sinθ/cosθ. From this, we can set up the following equation:
(3/4) = sinθ/cosθ
To solve for sinθ, multiply both sides of the equation by cosθ:
(3/4)cosθ = sinθ
Now, let's use the Pythagorean identity to find the value of cosθ.
sin^2θ + cos^2θ = 1
Rearranging the equation, we have:
cos^2θ = 1 - sin^2θ
cosθ = ±√(1 - sin^2θ)
Since θ is in the second quadrant, we know that cosθ is negative. Therefore, we take the negative value of cosθ:
cosθ = -√(1 - sin^2θ)
Now, substitute the value of sinθ from the earlier equation into the expression for cosθ:
cosθ = -√(1 - (3/4)^2)
cosθ = -√(1 - 9/16)
cosθ = -√(16/16 - 9/16)
cosθ = -√(7/16)
cosθ = -√7/4
Now, we can substitute these values into the double-angle identity for sine:
sin2θ = 2sinθcosθ
sin2θ = 2 * (3/4) * (-√7/4)
sin2θ = -3√7/8
Therefore, sin2θ is equal to -3√7/8.