done many times before.
Look at Related Questions below, here is one of them
http://www.jiskha.com/display.cgi?id=1491961364
sin(θ − ϕ);
tan(θ) = 5/12
θ in Quadrant III,
sin(ϕ) = − sqaure root 10/10 ϕ in Quadrant IV
I tried this numerous times can somone please help me
5 answers
The answer I keep getting is -3 square root of 10 over 130
so, why don't you show us how you got it? I get
in QIII
tanθ = sinθ/cosθ = 5/12, so
sinθ = -5/13
cosθ = -12/13
in QIV
sinϕ = -1/√10
cosϕ = 3/√10
You don't say what you want to do with that, but
sin(θ−ϕ) = (-5/13)(3/√10)-(-12/13)(-1/√10) = 21/(13√10)
in QIII
tanθ = sinθ/cosθ = 5/12, so
sinθ = -5/13
cosθ = -12/13
in QIV
sinϕ = -1/√10
cosϕ = 3/√10
You don't say what you want to do with that, but
sin(θ−ϕ) = (-5/13)(3/√10)-(-12/13)(-1/√10) = 21/(13√10)
Thats still the wrong answer
-27/13(square root 10)
1 answer