sin(θ − ϕ);
tan(θ) = 3/4,
θ in Quadrant III,
sin(ϕ) = − (3sqrt 10)/10,
ϕ in Quadrant IV
To find the value of sin(θ - ϕ), we need to use some trigonometric identities and the given information about the values of θ and ϕ.
First, let's find the values of sin(θ) and cos(θ) using the given information. Since tan(θ) = 3/4 and θ is in Quadrant III, we know that θ is negative because the tangent is negative in Quadrant III. Using the Pythagorean identity, we can find the value of cos(θ):
cos(θ) = ±sqrt(1 - tan^2(θ))
cos(θ) = ±sqrt(1 - (3/4)^2)
cos(θ) = ±sqrt(1 - 9/16)
cos(θ) = ±sqrt(16/16 - 9/16)
cos(θ) = ±sqrt(7/16)
Since cosine is positive in Quadrant III, we take the positive value of cos(θ):
cos(θ) = sqrt(7)/4
Now, let's find the values of sin(ϕ) and cos(ϕ) using the given information. Since sin(ϕ) = - (3sqrt(10))/10 and ϕ is in Quadrant IV, we know that both sin(ϕ) and cos(ϕ) are positive in Quadrant IV.
sin(ϕ) = - (3sqrt(10))/10
Now that we have the values of sin(θ), cos(θ), sin(ϕ), and cos(ϕ), we can use the trigonometric identity for sin(θ - ϕ):
sin(θ - ϕ) = sin(θ) * cos(ϕ) - cos(θ) * sin(ϕ)
Substituting the known values:
sin(θ - ϕ) = sin(θ) * cos(ϕ) - cos(θ) * sin(ϕ)
sin(θ - ϕ) = (sqrt(7)/4) * (3sqrt(10))/10 - (3/4) * (- (3sqrt(10))/10)
sin(θ - ϕ) = (3sqrt(70))/(4 * 10) - (9sqrt(10))/(4 * 10)
sin(θ - ϕ) = (3sqrt(70) - 9sqrt(10))/(40)
Therefore, sin(θ - ϕ) = (3sqrt(70) - 9sqrt(10))/(40).
To find the value of sin(θ - ϕ), we can use the trigonometric identity sin(A - B) = sin(A)cos(B) - cos(A)sin(B).
Given that tan(θ) = 3/4 and θ is in Quadrant III, we can determine the values of sin(θ) and cos(θ) using the Pythagorean identity.
Since tan(θ) = sin(θ)/cos(θ), we have sin(θ)/cos(θ) = 3/4.
Cross-multiplying, we get 4sin(θ) = 3cos(θ).
Using the Pythagorean identity sin^2(θ) + cos^2(θ) = 1, we have (3cos(θ))^2 + cos^2(θ) = 16(sin^2(θ) + cos^2(θ)).
Expanding, we get 9cos^2(θ) + cos^2(θ) = 16.
Combining like terms, we have 10cos^2(θ) = 16.
Dividing by 10, we get cos^2(θ) = 16/10.
Taking the square root, we have cos(θ) = ± √(16/10) = ± √(8/5) = ± (2√2)/√5.
Since θ is in Quadrant III, sin(θ) is negative. Therefore, sin(θ) = -√(1 - cos^2(θ)) = -√(1 - (8/5) * (2/5)) = -√(1 - 16/25) = -√(9/25) = -3/5.
Given that sin(ϕ) = - (3√10)/10 and ϕ is in Quadrant IV, we know that sin(ϕ) is negative. Therefore, we have sin(ϕ) = - |sin(ϕ)| = -√(1 - cos^2(ϕ)).
Squaring both sides, we have (-sin(ϕ))^2 = 1 - cos^2(ϕ).
Substituting the value of sin(ϕ), we get (3√10/10)^2 = 1 - cos^2(ϕ).
Simplifying, we have 9/10 = 1 - cos^2(ϕ).
Rearranging, we have cos^2(ϕ) = 1 - 9/10 = 10/10 - 9/10 = 1/10.
Taking the square root, we have cos(ϕ) = ± √(1/10) = ±√10/10 = ±(√10)/10.
Now we can substitute the values of sin(θ), sin(ϕ), cos(θ), and cos(ϕ) into the expression sin(θ - ϕ) = sin(θ)cos(ϕ) - cos(θ)sin(ϕ):
sin(θ - ϕ) = (-3/5)(√10/10) - ((±(2√2)/√5)(-3√10)/10)
= -(3√10)/5 * √10/10 + (±(2√2)/√5)(3√10)/10
= -(3√10)^2 / (5 * 10) ± (2√2 * 3√10) / (10√5)
= - 9/50 + (2 * 3 * √(2 * 10)) / (10 * √5)
= - 9/50 + 6√20 / (10√5)
= - 9/50 + (6/10)(√20)/(√5)
= - 9/50 + (3/5)(√20)/(√5)
= - 9/50 + 3√4/(5)
= - 9/50 + 3 * 2 / 5
= - 9/50 + 6/5
= (- 9 + 60) / 50
= 51 / 50
Therefore, sin(θ - ϕ) = 51/50.
since tanθ = 3/4 in QIII,
sinθ = -3/5
cosθ = -4/5
since sinϕ = -3/√10 in QIV,
cosϕ = 1/√10
Now you have everything you need to evaluate sin(θ − ϕ)
using your formula.