Find each value if sin A = 12/13 with 90 degrees <= A <= 270 degrees and if sin B = -7/25 with -90 degrees <= B <= 90 degrees.
sin(A - B)
sin(A + B)
To find the value of sin(A - B) and sin(A + B), we can use the trigonometric identities for the sum and difference of angles.
1. Finding sin(A - B):
Using the identity sin(A - B) = sinA*cosB - cosA*sinB
First, we need to find the values of sinA, cosA, sinB, and cosB.
Given: sinA = 12/13 and -90 degrees <= A <= 270 degrees
To find cosA, we can use the Pythagorean identity sin^2(A) + cos^2(A) = 1. Since sinA = 12/13, we can rearrange the equation to solve for cosA:
cos^2(A) = 1 - sin^2(A)
cos^2(A) = 1 - (12/13)^2
cos^2(A) = 1 - 144/169
cos^2(A) = 25/169
Taking the square root of both sides, we get:
cosA = ± √(25/169)
cosA = ± 5/13
Since A is between 90 degrees and 270 degrees, cosA will be negative. Therefore, cosA = -5/13.
Similarly, we can find sinB and cosB using the given values:
sinB = -7/25
cosB = ± √(1 - sin^2(B))
cosB = ± √(1 - (-7/25)^2)
cosB = ± √(1 - 49/625)
cosB = ± √(576/625)
cosB = ± 24/25
Since B is between -90 degrees and 90 degrees, cosB will be positive. Therefore, cosB = 24/25.
Now, substitute the values in the formula for sin(A - B):
sin(A - B) = sinA*cosB - cosA*sinB
sin(A - B) = (12/13)*(24/25) - (-5/13)*(-7/25)
sin(A - B) = 288/325 + 35/325
sin(A - B) = 323/325
So, sin(A - B) = 323/325.
2. Finding sin(A + B):
Using the identity sin(A + B) = sinA*cosB + cosA*sinB
Substitute the values for sinA, cosA, sinB, and cosB into the formula for sin(A + B):
sin(A + B) = sinA*cosB + cosA*sinB
sin(A + B) = (12/13)*(24/25) + (-5/13)*(-7/25)
sin(A + B) = 288/325 + 35/325
sin(A + B) = 323/325
So, sin(A + B) = 323/325.
if sinA = 12/13, and since the since is +
A must be in II
using our standard 5-12-13 triangle,
cosA = -5/13
if sinB = -7/25,
to state -90 ≤ B ≤ 90 is ambiguous, since you could be on the right or the left of the y-axis
I will assume B is in III , (could be in IV)
then using our standard 7-24-25 right-angled triangle, we can say that
cosB = -24/25
ok, now
sin(a-b) = sinAcosB - cosAsinB
= (12/13)(-24/25) - (-5/13)(-7/25)
= (-288/325 - 35/325) = -323/325
Using the expansion for sin(A+B) , I will let you do the other one