Question

If sin A=2/3 cos B=3/4, angle A is in quadrant 2 and angle B is in quadrant 4. How do you evaluate sin(A-B) without find A and B?

Answer 1

Please see the explanation.

Explanation:

Given: `sin(A) = 2/3`

We will need the value of `cos(A)` so use the identity:

`cos(A) = +-sqrt(1 - sin^2(A))`

We are given that A is in the 2nd quadrant, therefore, we chose the negative value for the cosine:

`cos(A) = -sqrt(1 - sin^2(A))`

`cos(A) = -sqrt(1 - (2/3)^2)`

`cos(A) = -sqrt(1 - 4/9)`

`cos(A) = -sqrt(5/9)`

`cos(A) = -sqrt(5)/3`

Given: `cos(B) = 3/4`

We will need the `sin(B)` so use the identity:

`sin(B) = +-sqrt(1 - cos^2(B))`

We are given that B is in the 4th quadrant, therefore, we chose the negative value for the sine:

`sin(B) = -sqrt(1 - cos^2(B))`

`sin(B) = -sqrt(1 - (3/4)^2)`

`sin(B) = -sqrt(1 - 9/16)`

`sin(B) = -sqrt(7/16)`

`sin(B) = -sqrt(7)/4`

We have all of the values that we need to use the identity:

`sin(A - B) = sin(A)cos(B) - cos(A)sin(B)`

`sin(A - B) = (2/3)(3/4) - (-sqrt(5)/3)(-sqrt(7)/4)`

`sin(A - B) = (6/12) - (sqrt(35)/12)`

`sin(A - B) = (6 - sqrt(35))/12`

Answer 2

`sin(A-B)=(6-sqrt35)/12=1/2-sqrt35/12`.

Explanation:

The easiest way I can think of is to use the Pythagorean Identity to first find both `cosA` and `sinB`, then use the sum/difference angle identity for sine.

The trigonometric Pythagorean identity tells us that

`sin^2theta+cos^2theta=1`

which follows directly from the geometric identity for right triangles and the fact that `sintheta=y/r` and `costheta=x/r`. If `r` is the distance `(x,y)` is from the origin, then

`x^2+y^2=r^2`
`(x/r)^2+(y/r)^2=1` (divide both sides by `r^2`)
`cos^2theta+sin^2theta=1`

Anyway, if we know `sinA` (and which quadrant it ends in), we can solve for `cosA`, and vice versa:

`sin^2A+cos^2A=1`
`(2/3)^2+cos^2A=1`
`=>cosA=+-sqrt(1-(2/3)^2)=+-sqrt(5)/3`

Because we know `angleA` ends in `Q_"II"`, we know `cosA` must be negative, because `costheta` is always negative for `theta in Q_"II"`. Therefore,

`cosA=-sqrt5/3`

Similar reasoning can be done with `cosB=3/4` to show that `sinB=-sqrt7/4`.

Now that we have all the necessary elements, we can use the sum/difference angle identity:

`sin(A+-B)=sinAcosB+-cosAsinB`

`sin(A-B)=sinAcosB-cosAsinB`
`sin(A-B)=(2/3)(3/4)-((-sqrt5)/3)((-sqrt7)/4)`
`sin(A-B)=6/12-(sqrt35)/12`

`sin(A-B)=(6-sqrt35)/12=1/2-sqrt35/12`

Hope this helps!