Suppose sinA = 12/13 with 90º≤A≤180º. Suppose also that sinB = -7/25 with -90º≤B≤0º. Find cos(A - B).
To find cos(A - B), we can use the trigonometric identity:
cos(A - B) = cosA * cosB + sinA * sinB
First, let's find cosA and cosB using the given information.
Given sinA = 12/13, we can use the Pythagorean identity to find cosA:
cosA = √(1 - sin^2 A) = √(1 - (12/13)^2) = √(1 - 144/169) = √(25/169) = 5/13
Given sinB = -7/25, we can again use the Pythagorean identity to find cosB:
cosB = √(1 - sin^2 B) = √(1 - (-7/25)^2) = √(1 - 49/625) = √(576/625) = 24/25
Now, substitute the values of cosA and cosB into the formula for cos(A - B):
cos(A - B) = (5/13)(24/25) + (12/13)(-7/25)
= 120/325 - 84/325
= 36/325
Therefore, cos(A - B) = 36/325.
To find cos(A - B), we can use the trigonometric identity:
cos(A - B) = cosA * cosB + sinA * sinB
Given that sinA = 12/13 and sinB = -7/25, we need to find the values of cosA and cosB.
To find cosA, we can use the Pythagorean identity:
sin²A + cos²A = 1
Given that sinA = 12/13, we can find cosA as follows:
cos²A = 1 - sin²A
cos²A = 1 - (12/13)²
cos²A = 1 - 144/169
cos²A = (169 - 144)/169
cos²A = 25/169
cosA = ±√(25/169)
cosA = ±5/13
Since A is an angle in the second quadrant (90° ≤ A ≤ 180°), cosA must be negative. So we have:
cosA = -5/13
Similarly, to find cosB, we can use the Pythagorean identity:
sin²B + cos²B = 1
Given that sinB = -7/25, we can find cosB as follows:
cos²B = 1 - sin²B
cos²B = 1 - (-7/25)²
cos²B = 1 - 49/625
cos²B = (625 - 49)/625
cos²B = 576/625
cosB = ±√(576/625)
cosB = ±24/25
Since B is an angle in the fourth quadrant (-90° ≤ B ≤ 0°), cosB must be positive. So we have:
cosB = 24/25
Now, substituting the values of sinA, sinB, cosA, and cosB into the formula for cos(A - B), we get:
cos(A - B) = cosA * cosB + sinA * sinB
cos(A - B) = (-5/13) * (24/25) + (12/13) * (-7/25)
cos(A - B) = -120/325 + (-84/325)
cos(A - B) = (-120 - 84)/325
cos(A - B) = -204/325
Therefore, cos(A - B) = -204/325.
A is in QII, so cosA = -5/13
B is in QIV, so cosB = 24/25
Now, you have
cos(A-B) = cosAcosB + sinAsinB
...
What kinda answer is that Steve
EH?