sin(θ − ϕ); tan(θ) = 5/12 θ in Quadrant III, sin(ϕ) = − sqaure root10/10 ϕ in Quadrant IV.
I used the sin equation
sin(a)cos(b)-sin(a)cos(b)
However I am still getting the wrong answer
in QIII
sinθ = -5/13
cosθ = -12/13
in QIV
sinϕ = -1/√10
cosϕ = 3/√10
sin(θ − ϕ) = (-5/13)(3/√10)-(-12/13)(3/√10) = 21/(13√10)
it was incorrect again :(
Better check my math, then. Also, maybe they don't want the radical in the denominator.
To correctly evaluate the expression sin(θ - ϕ), you can use the identity sin(a - b) = sin(a)cos(b) - cos(a)sin(b). In this case, let's label θ as a and ϕ as b.
Given that tan(θ) = 5/12 in Quadrant III, we can determine the values of sin(θ) and cos(θ) using the Pythagorean identity:
sin^2(θ) + cos^2(θ) = 1
Since we are in Quadrant III and tan(θ) = 5/12, we know that sin(θ) will be negative and cos(θ) will be negative.
We can let sin(θ) = -5, and cos(θ) = -12 (up to the appropriate scaling factor).
Next, given that sin(ϕ) = -√10/10 in Quadrant IV, we can determine the values of sin(ϕ) and cos(ϕ) using the Pythagorean identity in a similar way.
Let sin(ϕ) = -√10, and cos(ϕ) = 10 (up to the appropriate scaling factor).
Now substituting these values into the identity sin(a - b) = sin(a)cos(b) - cos(a)sin(b), we have:
sin(θ - ϕ) = sin(θ)cos(ϕ) - cos(θ)sin(ϕ)
= (-5)(10) - (-12)(-√10)
= -50 + 12√10
Therefore, sin(θ - ϕ) = -50 + 12√10.
Make sure to double-check the values you plugged in for sin(θ) and cos(θ), as well as sin(ϕ) and cos(ϕ), to ensure accuracy in the calculations.