Given that tan¤=3/4 and ¤ is in the second quadrant find sin2¤ (¤=theta)

see other post.

cos theta = + OR - 1 / sqrt ( 1 + tan ^ 2 theta )

In Quadrant II, cosine are negative so :

cos theta = - 1 / sqrt ( 1 + tan ^ 2 theta )

cos theta = - 1 / sqrt [ 1 + ( 3 / 4 ) ^ 2 theta ]

cos theta = - 1 / sqrt ( 1 + 9 / 16 )

cos theta = - 1 / sqrt ( 16 / 16 + 9 / 16 )

cos theta = - 1 / sqrt ( 25 / 16 )

cos theta = - 1 / ( 5 / 4 )

cos theta = - 4 / 5

sin ^ 2 theta + cos ^ 2 theta = 1

sin ^ 2 theta = 1 - cos ^ 2 theta

sin theta = + OR - sqrt ( 1 - cos ^ 2 theta )

In Quadrant II, sine are positive so :

sin theta = sqrt ( 1 - cos ^ 2 theta )

sin theta = sqrt [ 1 - ( - 4 / 5 ) ^ 2 ]

sin theta = sqrt ( 1 - 16 / 25 )

sin theta = sqrt ( 25 / 25 - 16 / 25 )

sin theta = sqrt ( 9 / 25 )

sin theta = 3 / 5

sin 2 theta = 2 sin theta cos theta

sin 2 theta = 2 * 3 / 5 * - 4 / 5

sin 2 theta = - 24 / 25

To find sin(2θ), we can use the double angle identity for sine, which states that sin(2θ) = 2sin(θ)cos(θ).

From the given information that tan(θ) = 3/4, we can find the values of sine and cosine using the Pythagorean identity for tangent: tan(θ) = sin(θ)/cos(θ).

Since θ is in the second quadrant, which is defined by negative values for x and positive values for y, we can determine that sin(θ) is positive and cos(θ) is negative.

Given tan(θ) = 3/4, we can set up the following equation:
sin(θ) / cos(θ) = 3/4.

Cross-multiplying, we have:
4sin(θ) = 3cos(θ).

Now, to find sin(θ) and cos(θ), we need to find the value of the hypotenuse of the triangle formed in the second quadrant. We can use the Pythagorean theorem, which states that for any right triangle, the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b).

Let's assume the opposite side to be 3x and the adjacent side to be 4x (in order to match the given information from the tangent). If we label the hypotenuse as c, we can set up the following equation:
(3x)^2 + (4x)^2 = c^2.

Simplifying the equation, we get:
9x^2 + 16x^2 = c^2.
25x^2 = c^2.
c = 5x.

Now, we have the values of sin(θ), cos(θ), and the hypotenuse (c):
sin(θ) = 3x/c = 3x/5x = 3/5.
cos(θ) = -4x/c = -4x/5x = -4/5.

Next, we can calculate sin(2θ) using the double angle identity for sine:
sin(2θ) = 2sin(θ)cos(θ).

Substituting the values we found earlier:
sin(2θ) = 2 * (3/5) * (-4/5).

Simplifying the expression, we get:
sin(2θ) = -24/25.

Therefore, sin²(θ) = (-24/25)² = 576/625.

Hence, the value of sin²(θ) is 576/625.