(Updated )Evaluate the expression under the given conditions sin(θ − ϕ); tan(θ) = ¾, θ in Quadrant III, sin(ϕ) = − 3 sqroot10/10, ϕ in Quadrant IV

Question

I need help with this question. Please include steps. Thank you :)

Answer 1

tanθ=.75

θ=36.8698...........=216.8698..............quad3

sin(ϕ) = − 3

" =-1.5707...........quad4

1^.5=1

sin(θ − ϕ)

" (216.87-(-1.57))=

" 218.44=-.622

Answer 2

First lets look at sin(ϕ) = − 3√10/10

Remember that sin(x) = opposite/hypotenuse so the opposite side must be -3√10 and the hypotenuse must be 10. Using the Pythagorean Theorem, we can solve for the side in the "x" direction.

Now lets look at tan(θ) = 3/4. Remember that tan(x) = opposite/adjacent so the opposite side must be 3 and the adjacent side must be 4. But then, looking at the restriction that θ is in Q3, we realize that the opposite side must be -3 and the adjacent side must be -4. Using the Pythagorean Theorem, we can solve for the hypotenuse like this:

Now, lets look at sin(θ - Φ)

Using the angle addition/subtraction identity, we know that sin(x – y) = sin(x)cos(y) – cos(x)sin(y)

so, in this case, sin(θ - Φ) = sin(θ)cos(Φ) - cos(θ)sin(Φ)

Using the diagrams, we can identify each of those pieces.

sin(θ) = -3/5

cos(Φ) = √10/10

cos(θ) = -4/5

sin(Φ) = -3√10/10

So:

sin(θ - Φ) = (-3/5)(√10/10) - (-4/5)(-3√10/10)

sin(θ - Φ) = -3√10/50 - 12√10/50

sin(θ - Φ) = -15√10/50

sin(θ - Φ) = -3√10/10

A strange answer considering it's also the sine of Φ but that's the way it works sometimes.