Evaluate the expression under the given conditions:

Question

sin(θ+∅); sin(θ)=4/5, θ in quadrant I, cos(∅)=-(2√5)/5, ∅ in quadrant II

Answer 1

We know the identity sin(a+b) = sin(a)cos(b) + sin(b)cos(a) --> sin(θ+ø) = sin(θ)cos(∅) + sin(∅)cos(θ)

We can substitute the given values for theta and phi into the identity:

For the first term, we substitute for sin(θ)cos(∅): (4/5)(-2√5/5) = -8√5/25

For the second term, we need to determine sin(∅) and cos(θ). Lets start with cos(θ). We know that the 3-4-5 triangle is a special 30-60-90 triangle. Therefore, cos(θ) will be the ratio of the remaining side of the triangle--> cos(θ) = 3/5. There is no negative sign because θ is in quadrant I, and cosine in the first quadrant is always positive.

Now we need to determine sin(∅). We need to figure out the remaining side of this triangle. We can use Pythagorean's Theorem to do so. Since cos(∅) = -(2√5)/5, we know that the hypotenuse is 5, and one side of the triangle is 2√5. So we get (2√5)^2 + x^2 = 5^2 --> 20 + x^2 = 25 --> x^2 = 5 --> x = √5. Therefore, the remaining side is equal to √5, and sin(∅) = √5/5. The sine function is positive in quadrant II.

Now we can figure out the second term. sin(∅)cos(θ) = (√5/5)(3/5) = 3√5/25.

Now we can add this term to the previous term we found and get our answer: sin(θ+ø) = -8√5/25 + 3√5/25 = -5√5/25 = -√5/5.

Answer 2

sin(θ + ∅) = sin θ cos ∅ + cos θ sin ∅

In quadrant 1, sin and cos are positive, so if sin θ = 4/5, cos θ = √(1 - (4/5)²) = √(1 - (16/25)) = √(9/25) = 3/5

(Alternatively, we could just recall the 3-4-5 right triangle).

In quadrant 2, cos is negative and sin is positive, so sin ∅ = √(1 - (-2√5/5)²) = √(1 - 20/25) = √(1/5) = √5 / 55

(Alternatively, we could recall the 1-2-√5 triangle).

Then, sin(θ + ∅) = 4/5 * -2√5/5 + 3/5 * 1√5/5 = -5√5/25 = -√5/5