tan(θ + ϕ); cos(θ) = − 1/3
θ in Quadrant III, sin(ϕ) = 1/4 ϕ in Quadrant II
I am really struggling with this idk why I keep getting the wrong answers
3 answers
me either. why don't you show your work? I've done several already; your turn. I'm sure we can pinpoint where you go astray.
i was given this equation Tan(a+b)=sin(a+b)/cos(a+b)
So I did (square root 8/3)(square root 15/4)+sin(1/4)cos(-1/3) divide that by (-1/3)(square root 15 over 4) - sin( square root 8 over 3)sin(1/4)
So I did (square root 8/3)(square root 15/4)+sin(1/4)cos(-1/3) divide that by (-1/3)(square root 15 over 4) - sin( square root 8 over 3)sin(1/4)
what are all those sin() stuff still hanging around? The final expression should just be a bunch of fractions.
tan(θ + ϕ); cos(θ) = − 1/3
θ in Quadrant III, sin(ϕ) = 1/4 ϕ in Quadrant II
You really need to review the signs of the trig functions in the various quadrants.
In QIII,
sinθ = -√8/3
cosθ = -1/3
In QII,
sinϕ = 1/4
cosϕ = -√15/4
tan(θ+ϕ) = sin(θ+ϕ)/cos(θ+ϕ)
= (sinθcosϕ+cosθsinϕ)/(cosθcosϕ-sinθsinϕ)
= ((-√8/3)(-√15/4)+(-1/3)(1/4))/((-1/3)(-√15/4)-(-√8/3)(1/4))
= (32√2-9√15)/7
Or, using tan(θ+ϕ) = (tanθ+tanϕ)/(1-tanθ*tanϕ)
tanθ = sinθ/cosθ = √8
tanϕ = sinϕ/cosϕ = -1/√15
tan(θ+ϕ) = (√8 - 1/√15)/(1+√8/√15) = (32√2-9√15)/7
tan(θ + ϕ); cos(θ) = − 1/3
θ in Quadrant III, sin(ϕ) = 1/4 ϕ in Quadrant II
You really need to review the signs of the trig functions in the various quadrants.
In QIII,
sinθ = -√8/3
cosθ = -1/3
In QII,
sinϕ = 1/4
cosϕ = -√15/4
tan(θ+ϕ) = sin(θ+ϕ)/cos(θ+ϕ)
= (sinθcosϕ+cosθsinϕ)/(cosθcosϕ-sinθsinϕ)
= ((-√8/3)(-√15/4)+(-1/3)(1/4))/((-1/3)(-√15/4)-(-√8/3)(1/4))
= (32√2-9√15)/7
Or, using tan(θ+ϕ) = (tanθ+tanϕ)/(1-tanθ*tanϕ)
tanθ = sinθ/cosθ = √8
tanϕ = sinϕ/cosϕ = -1/√15
tan(θ+ϕ) = (√8 - 1/√15)/(1+√8/√15) = (32√2-9√15)/7
3 answers