Given Tan(A) = 5 in Quadrant III and Sin(B) = ⅔ in Quadrant II, find Cos(A-B).
Steve had already given you a good hint how to do these types in your previous trig question, this is the same type.
You have to know the definitions of the trig functions in terms of x, y, and r and you have to know the CAST rule
tanA = 5/1 = y/x , but we are in III, so y = -5, x = -1
so r^2 = (-5)^2 + (-1)^2 = 26
r = √26
then sinA = -5/√26 , cosA = -1/√26
similarely if sin B = 2/3, and B is in II
cos B = -√5/3
cos(A-B) = cosAcosB + sinAsinB
= (-1/√26)(-√5/3) + (-5/√26)(2/3)
= (√5 - 10)/(3√26)
To find cos(A-B), we can use the following trigonometric identity:
cos(A-B) = cosA * cosB + sinA * sinB
First, let's find the values of cosA and sinA.
Since tan(A) = 5 in Quadrant III, we can determine that tan(A) = -5 (since the tangent is negative in Quadrant III).
Using the Pythagorean identity, we have:
tan(A) = sin(A) / cos(A)
-5 = sin(A) / cos(A)
Next, let's find the values of cosB and sinB.
Since sin(B) = ⅔ in Quadrant II, we can determine that sin(B) = √(1 - cos^2B).
By substituting the given value, we get:
2/3 = √(1 - cos^2B)
Now, we can solve for each angle using the information given.
For angle A:
tan(A) = -5, cos(A) = ?
Given that tan(A) = sin(A) / cos(A), we can substitute in the values:
-5 = sin(A) / cos(A)
sin(A) = -5cos(A)
Using the Pythagorean identity: sin^2(A) + cos^2(A) = 1
We can substitute the relation sin(A) = -5cos(A):
(-5cos(A))^2 + cos^2(A) = 1
25cos^2(A) + cos^2(A) = 1
26cos^2(A) = 1
cos^2(A) = 1/26
cos(A) = ± √(1/26)
Since we are in Quadrant III, cos(A) is negative, so:
cos(A) = -√(1/26)
For angle B:
sin(B) = 2/3, cos(B) = ?
By using the Pythagorean identity, we can find cos(B):
sin^2(B) + cos^2(B) = 1
(2/3)^2 + cos^2(B) = 1
4/9 + cos^2(B) = 1
cos^2(B) = 5/9
cos(B) = ±√(5/9)
We are in Quadrant II, so cos(B) is negative:
cos(B) = -√(5/9)
Now, we can substitute the values into the formula for cos(A-B):
cos(A-B) = cosA * cosB + sinA * sinB
cos(A-B) = (-√(1/26)) * (-√(5/9)) + (-5√(1/26)) * (2/3)
Simplifying this expression, we get:
cos(A-B) = (√(5/234)) + (-10√(1/234))
cos(A-B) = √(5/234) - 10√(1/234)
Therefore, the value of cos(A-B) is √(5/234) - 10√(1/234).
To find Cos(A-B), we need to know the values of Cos(A) and Sin(B) in order to evaluate the cosine of the difference of two angles.
To find Cos(A), we can use the identity Cos^2(A) + Sin^2(A) = 1. Since we know that Tan(A) = 5, and Tan(A) = Sin(A)/Cos(A), we can solve for Sin(A) and use the identity to find Cos(A).
Tan(A) = 5
Sin(A)/Cos(A) = 5
Sin(A) = 5 * Cos(A)
Using the identity Cos^2(A) + Sin^2(A) = 1:
Cos^2(A) + (5 * Cos(A))^2 = 1
Cos^2(A) + 25 * Cos^2(A) = 1
26 * Cos^2(A) = 1
Cos^2(A) = 1/26
Since we know that A is in Quadrant III, where Cos(A) is negative, we take the negative square root:
Cos(A) = -1/√26
To find Sin(B), we know that Sin(B) = ⅔. Since B is in Quadrant II, where Sin(B) is positive, we can use this value directly.
Sin(B) = ⅔
Now we have the values of Cos(A) and Sin(B), and we can evaluate Cos(A-B) using the formula:
Cos(A-B) = Cos(A) * Cos(B) + Sin(A) * Sin(B)
Substituting the values we found, we have:
Cos(A-B) = (-1/√26) * Cos(B) + (5 * (-1/√26)) * (⅔)
Without knowing the value of Cos(B), we can't simplify this any further.