sin(θ + ϕ);
sin(θ) = 5/13 θ in Quadrant I,
cos(ϕ) = − 2root 5 over 5 ϕ in Quadrant II
please give final answer this is bc i have a lot of trouble simplifying the final answer thank you
Well, luckily for you, I'm the master of simplifying things and making them fun! Let's hop right into it!
We're given that sin(θ) = 5/13, which means that in Quadrant I, the opposite side of the triangle is 5, and the hypotenuse is 13.
On the other hand, we're told that cos(ϕ) = -2√5/5, which means that in Quadrant II, the adjacent side is -2√5, and the hypotenuse is 5.
Now, let's find sin(θ + ϕ) using the angle addition formula for sine:
sin(θ + ϕ) = sin(θ)cos(ϕ) + cos(θ)sin(ϕ)
Since we already know the values of sin(θ) and cos(ϕ), we can plug them in:
sin(θ + ϕ) = (5/13)(-2√5/5) + cos(θ)sin(ϕ)
To simplify, let's cancel out common factors:
sin(θ + ϕ) = -10√5/65 + cos(θ)sin(ϕ)
Now, here comes the tricky part. We need to find the values of cos(θ) and sin(ϕ) to get the final answer. Unfortunately, you didn't provide those values, so I can't give you the exact answer. But hey, at least we had some fun along the way, right? If you provide the values, I'll be more than happy to help you simplify the final answer!
To find sin(θ + ϕ), we'll use the trigonometric identity:
sin(θ + ϕ) = sin(θ)cos(ϕ) + cos(θ)sin(ϕ)
Given that sin(θ) = 5/13 (with θ in Quadrant I) and cos(ϕ) = -2√5/5 (with ϕ in Quadrant II), we can substitute the values into the identity.
sin(θ + ϕ) = (5/13)(-2√5/5) + [√(1 - (5/13)^2)][(2√5)/5]
First, let's simplify each term:
(5/13)(-2√5/5) = -10√5/65
[√(1 - (5/13)^2)][(2√5)/5] = [√(1 - 25/169)][(2√5)/5] = [(√(144/169)][(2√5)/5] = [12/13][(2√5)/5] = 24√5/65
Now, add the two terms:
sin(θ + ϕ) = -10√5/65 + 24√5/65
To add these fractions, we need a common denominator of 65:
sin(θ + ϕ) = (-10√5 + 24√5)/65
Combining like terms:
sin(θ + ϕ) = 14√5/65
So, the final answer is sin(θ + ϕ) = 14√5/65.
To simplify the expression sin(θ + ϕ), we can use the angle addition formula for sine:
sin(θ + ϕ) = sin(θ)cos(ϕ) + cos(θ)sin(ϕ)
Given that sin(θ) = 5/13, we can determine the values of cos(θ) and sin(ϕ) using the Pythagorean identity:
cos²(θ) + sin²(θ) = 1
Since θ is in Quadrant I, cos(θ) will be positive:
cos(θ) = sqrt(1 - sin²(θ))
cos(θ) = sqrt(1 - (5/13)²)
cos(θ) = sqrt(1 - 25/169)
cos(θ) = sqrt(144/169)
cos(θ) = 12/13
Next, since ϕ is in Quadrant II, sin(ϕ) will be positive:
sin(ϕ) = sqrt(1 - cos²(ϕ))
sin(ϕ) = sqrt(1 - (-2sqrt(5)/5)²)
sin(ϕ) = sqrt(1 - 20/25)
sin(ϕ) = sqrt(5/25)
sin(ϕ) = sqrt(1/5)
sin(ϕ) = 1/sqrt(5)
sin(ϕ) = sqrt(5)/5
Now we can substitute the values into the angle addition formula:
sin(θ + ϕ) = sin(θ)cos(ϕ) + cos(θ)sin(ϕ)
sin(θ + ϕ) = (5/13)(-2sqrt(5)/5) + (12/13)(sqrt(5)/5)
sin(θ + ϕ) = -10sqrt(5)/65 + 12sqrt(5)/65
sin(θ + ϕ) = 2sqrt(5)/65
Therefore, the final answer is sin(θ + ϕ) = 2sqrt(5)/65.
Funny - you and Ruth need to compare notes ...
in QI
sinθ = 5/13
cosθ = 12/13
in QII
sinϕ = 1/√5
cosϕ = -2/√5
sin(θ+ϕ) = (5/13)(-2/√5)+(12/13)(1/√5) = 2/(13√5)