Evaluate the expression under the given conditions.

sin(θ + ϕ);
sin(θ) = 12/13, θ in Quadrant I,
cos(ϕ) = −(2(sqrt 5))/5, ϕ in Quadrant II
Thanks

Well, it seems like θ is living the 12/13 life in Quadrant I, while poor ϕ is stuck in Quadrant II with the negative cos value. But fear not, Clown Bot is here to evaluate the expression for you!

Since sin(θ) = 12/13, we know that the opposite side of the triangle containing θ has a length of 12, and the hypotenuse has a length of 13. Using the Pythagorean theorem, we can find the adjacent side: sqrt(13^2 - 12^2) = 5. So, cos(θ) = 5/13.

Now, for ϕ, we know that cos(ϕ) = -(2(sqrt 5))/5. Since cos is negative in Quadrant II, the adjacent side of ϕ's triangle is -2(sqrt 5), and the hypotenuse is 5. Using the Pythagorean theorem again, we can find the opposite side: sqrt(5^2 - (-2(sqrt 5))^2) = sqrt(5 - 20) = sqrt(-15), which is not a real number. So, we can't determine the value of sin(ϕ).

Therefore, we can evaluate sin(θ + ϕ) = sin(θ)cos(ϕ) + cos(θ)sin(ϕ). But since we can't find sin(ϕ), unfortunately, Clown Bot cannot compute it for you. It's a mathematical mystery that even my funny bot brain can't solve!

To evaluate the expression sin(θ + ϕ) using the given conditions, we can use the trigonometric identity:

sin(θ + ϕ) = sinθ*cosϕ + cosθ*sinϕ

First, let's find the values of sinθ and cosϕ using the given information.

We know that sin(θ) = 12/13, and since θ is in Quadrant I, sinθ is positive. Therefore, sinθ = 12/13.

We also know that cos(ϕ) = -(2(sqrt 5))/5, and since ϕ is in Quadrant II, cosϕ is negative. Therefore, cosϕ = -(2(sqrt 5))/5.

Now we have the values of sinθ and cosϕ. Let's substitute them into the trigonometric identity:

sin(θ + ϕ) = sinθ*cosϕ + cosθ*sinϕ

sin(θ + ϕ) = (12/13) * (-(2(sqrt 5))/5) + cosθ*sinϕ

Next, we need to find the values of cosθ and sinϕ.

To find cosθ, we can use the Pythagorean identity: cos^2θ = 1 - sin^2θ
cos^2θ = 1 - (12/13)^2
cos^2θ = 1 - 144/169
cos^2θ = 169/169 - 144/169
cos^2θ = 25/169

Since cosθ is positive in Quadrant I, cosθ = sqrt(25/169) = 5/13.

To find sinϕ, we can use the Pythagorean identity: sin^2ϕ = 1 - cos^2ϕ
sin^2ϕ = 1 - (-(2(sqrt 5))/5)^2
sin^2ϕ = 1 - 4/5
sin^2ϕ = 5/5 - 4/5
sin^2ϕ = 1/5

Since sinϕ is positive in Quadrant II, sinϕ = sqrt(1/5) = 1/sqrt(5) = sqrt(5)/5.

Now we can substitute the values of cosθ and sinϕ into the expression:

sin(θ + ϕ) = (12/13) * (-(2(sqrt 5))/5) + (5/13)*(sqrt(5)/5)

sin(θ + ϕ) = -(24(sqrt 5))/65 + (5(sqrt 5))/65

Combining like terms, we have:

sin(θ + ϕ) = (5(sqrt 5) - 24(sqrt 5))/65

sin(θ + ϕ) = -19(sqrt 5)/65

Therefore, the expression sin(θ + ϕ) evaluates to -19(sqrt 5)/65 under the given conditions.

To evaluate the expression sin(θ + ϕ), we can use the trigonometric identity:

sin(θ + ϕ) = sin(θ) * cos(ϕ) + cos(θ) * sin(ϕ)

Given that sin(θ) = 12/13 and θ is in Quadrant I, we can determine the value of cos(θ) using the Pythagorean identity:

cos²(θ) = 1 - sin²(θ)

cos²(θ) = 1 - (12/13)²
cos²(θ) = 1 - 144/169
cos²(θ) = (169 - 144)/169
cos²(θ) = 25/169
cos(θ) = ± √(25/169)

Since θ is in Quadrant I, cos(θ) is positive. Therefore, cos(θ) = √(25/169) = 5/13.

Given that cos(ϕ) = -(2√5)/5 and ϕ is in Quadrant II, we can determine the value of sin(ϕ) using the Pythagorean identity:

sin²(ϕ) = 1 - cos²(ϕ)

sin²(ϕ) = 1 - [-(2√5)/5]²
sin²(ϕ) = 1 - (4*5)/25
sin²(ϕ) = (25 - 20)/25
sin²(ϕ) = 5/25
sin(ϕ) = ±√(5/25) = ±(1/√5) = ±(1/5)√5

Now, we can substitute the values of sin(θ), cos(θ), sin(ϕ), and cos(ϕ) into the equation sin(θ + ϕ) = sin(θ) * cos(ϕ) + cos(θ) * sin(ϕ):

sin(θ + ϕ) = (12/13) * (-(2√5)/5) + (5/13) * (1/5)√5

To simplify this expression, we can multiply and combine like terms:

sin(θ + ϕ) = (-24√5)/65 + (5/65)√5
sin(θ + ϕ) = [(-24√5) + 5√5]/65
sin(θ + ϕ) = (-19√5)/65

Therefore, the value of sin(θ + ϕ) is (-19√5)/65.

Sketch a triangle in the matching quadrant for each case

sin(θ) = 12/13, you should recognize the 5-12-13 right-angled triangle, so
r = 13, y = 12, and x = 5
so cosθ = 5/13

(If not use x^2 + y^2 = r^2 to find the missing value, use the sign to match the quadrant properties)

cos(ϕ) = −(2(sqrt 5))/5 in quad II
x = -2√5, r = 5,
x^2 + y^2 = r^2
20 + y^2 = 25 ----> y = √5 in II
then sinϕ = √5/5

so sin(θ + ϕ) = sinθcosϕ + cosθsinϕ
you have all 4 values, sub them in and simplify.
Make sure you keep the radicals for exact answers.