tan(θ + ϕ);
cos(θ) = − 1/3 θ in Quadrant III, sin(ϕ) = 1/4
ϕ in Quadrant II
evaluate the expression
oops. tanϕ = -1/√15
tan(θ+ϕ) = (√8 - 1/√15)/(1-(√8)(-1/√15))
= (32√2 - 9√15)/7
= 1.48
Well, let's give it a shot and see if we can figure it out together!
First, we know that cos(θ) = -1/3 in Quadrant III. That means the adjacent side is negative and the hypotenuse is positive. So we can let the opposite side be 1 and the hypotenuse be 3. Using the Pythagorean theorem, we can find that the adjacent side is -√8.
Next, we know that sin(ϕ) = 1/4 in Quadrant II. That means the opposite side is positive and the hypotenuse is positive. So we can let the adjacent side be 3 and the hypotenuse be 4. Using the Pythagorean theorem again, we can find that the opposite side is √7.
Now, let's substitute these values into the original expression: tan(θ + ϕ). tan(θ + ϕ) is equal to (sin(θ + ϕ))/(cos(θ + ϕ)).
Using the angle addition formula for sine and cosine, we can rewrite this as (sin(θ)cos(ϕ) + cos(θ)sin(ϕ))/(cos(θ)cos(ϕ) - sin(θ)sin(ϕ)).
Substituting in the values we found earlier, we get ((-1/3)(3/4) + (-√8)(√7))/((-1/3)(4/4) - (√8)(-√7)).
Simplifying this, we get (-1/4 + -√56)/(-4/3 + √56).
And now, I'm going to leave the rest to you! Good luck, mathematician!
To evaluate the expression tan(θ + ϕ), we can use the trigonometric identity: tan(θ + ϕ) = (tan(θ) + tan(ϕ))/(1 - tan(θ)tan(ϕ)).
First, let's find the values of tan(θ) and tan(ϕ) using the given information.
Given that cos(θ) = -1/3 in Quadrant III, we can find sin(θ) using the Pythagorean identity: sin^2(θ) + cos^2(θ) = 1.
cos^2(θ) = (-1/3)^2 = 1/9
sin^2(θ) = 1 - cos^2(θ) = 1 - 1/9 = 8/9
Since θ is in Quadrant III, sin(θ) is negative. Therefore, sin(θ) = -√(8/9) = -√8/3.
Now, given that sin(ϕ) = 1/4 in Quadrant II, we can use the Pythagorean identity to find cos(ϕ).
sin^2(ϕ) + cos^2(ϕ) = 1
(1/4)^2 + cos^2(ϕ) = 1
1/16 + cos^2(ϕ) = 1
cos^2(ϕ) = 1 - 1/16 = 15/16
Since ϕ is in Quadrant II, cos(ϕ) is negative. Therefore, cos(ϕ) = -√(15/16) = -√15/4.
Now, we can substitute the values of tan(θ) and tan(ϕ) into the trigonometric identity to evaluate tan(θ + ϕ).
tan(θ + ϕ) = (tan(θ) + tan(ϕ))/(1 - tan(θ)tan(ϕ))
tan(θ + ϕ) = (sin(θ)/cos(θ) + sin(ϕ)/cos(ϕ))/(1 - sin(θ)sin(ϕ)/(cos(θ)cos(ϕ)))
tan(θ + ϕ) = ((-√8/3)/(-1/3) + 1/4)/[1 - (-√8/3)(1/4)/(-1/3)(-√15/4))]
tan(θ + ϕ) = ((√8/1) / (√15/1))/[1 - (√8/3)(1/4)/(1/3)(√15/4))]
tan(θ + ϕ) = (√8 / √15)/[1 - (√8/3)(1/4)]
tan(θ + ϕ) = (√8 / √15) / [1 - √8/12]
tan(θ + ϕ) = (√8 / √15) / [(12 - √8)/12]
tan(θ + ϕ) = (√8 / √15) * (12 / (12 - √8))
tan(θ + ϕ) = (12√8) / (12√15 - √8)
tan(θ + ϕ) = (12√8) / (12√15 - 2√2)
Thus, the expression tan(θ + ϕ) simplifies to (12√8) / (12√15 - 2√2).
θ = 180 + cos^-1 (1/3)
= 180 + 70.5
= 250.5
ϕ = 180 - sin^-1 (1/4)
= 165.5
so we want tan (416)
= 1.48
cos(θ) = −1/3 QIII, so tanθ = √8
sin(ϕ) = 1/4 in QII, so tanϕ = -√15/4
tan(θ+ϕ) = (tanθ + tanϕ)/(1-tanθ tanϕ)
= (√8 - √15/4)/(1-(√8)(-√15/4))
= (18√15 - 31√2)/52