Using the identity sin^2(x) + cos^2(x) = 1, we have cos^2(x) = 1 - (3/5)^2 = 1 - 9/25 = 16/25. Therefore, cos(x) = ±4/5.

Using sin^2(α) + cos^2(α) = 1, we find cos(α) = √(1 - 0.6^2) = √(1 - 0.36) = √0.64 = 0.8.

The angles where sin(θ) = 0 in the interval [0, 2π] are θ = 0 and θ = π.

sin(θ) = 0 at θ = 0° and 180°.

sin(30°) = 0.5, so θ = 30°.

Using inverse sine, θ ≈ 36.87°

Using sin²(θ) + cos²(θ) = 1, cos²(θ) = 1 - 0.64 = 0.36, cos(θ) = 0.6

Using sin²(θ) + cos²(θ) = 1, cos(θ) = √(1 - 0.8²) = √(0.36) = 0.6.

sin(60°) = √3/2 ≈ 0.866, so θ = 60°.

sin(60°) = √3/2 ≈ 0.866, so θ = 60°.

sin(90°) = 1, so θ = 90°.

sin(90°) = 1, so θ = 90°.

sin(30°) = 1/2, so θ = 30°.

Using sin²(θ) + cos²(θ) = 1, cos(θ) = √(1 - (1/√2)²) = √(1/2) = √2/2

Using the identity sin^2(θ) + cos^2(θ) = 1, we have cos^2(θ) = 1 - (1/√2)^2 = 1 - 1/2 = 1/2. Thus, cos(θ) = ±1/√2.

sin(θ) = 1/√2 at θ = 45°.

sin(θ) = 1/√2 at θ = 45° and θ = 225°.

sin(θ) = 1/√2 at θ = 45°.

sin(45°) = 1/√2, so θ = 45°

Using the Pythagorean identity, cos(θ) = √(1 - sin²(θ)) = √(1 - (3/5)²) = 4/5.

Using the identity tan(θ) = sin(θ)/cos(θ) and cos(θ) = √(1 - sin^2(θ)), we find cos(θ) = 3/5. Thus, tan(θ) = (4/5)/(3/5) = 4/3.

The equation sin^(-1)(x) + cos^(-1)(x) = π/2 holds for all x in the domain of the functions, which is [-1, 1]. Therefore, x can be any value in this range.

Using the identity sin^(-1)(x) + cos^(-1)(x) = π/2, we can conclude that x can take any value in the range [-1, 1].

If sin^(-1)(x) = π/4, then x = sin(π/4) = √2/2.

Since some fruits are apples and all apples are sweet, it follows that some fruits are sweet.

V is the brother of T as both are children of U.

S is to the left of T.

Using the identity tan A = sin A / cos A, we can find sin A = tan A * cos A. Using the Pythagorean identity, we find sin A = 3/5.

Since tan θ = sin θ / cos θ and tan θ = 1, we have sin θ = cos θ. For θ = 45°, sin θ = 1/√2.

If tan(x) = 1, then sin(x) = cos(x). Therefore, sin(x) + cos(x) = 2sin(x) = 2(1/√2) = √2.