If sin A= 1/4 and sin B= -1/2, with A in QII and B in QIV, find tan(A+B)
in QII if sinA = 1/4, tanA = -1/√15
in QIV, if sinB = -1/2, tanB = -1/√3
tan(A+B) = (tanA+tanB)/(1-tanAtanB)
= (-1/√15 - 1/√3)/(1-1/√15√3)
= (√3+√15)/(√45-1)
Oops. That would be
-(√3+√15)/(√45-1)
To find tan(A+B), we need to know the values of both sin(A) and sin(B). Given that sin(A) = 1/4 and sin(B) = -1/2, we can use the trigonometric identity for tan(A+B) to calculate the value.
First, we need to find the values of cos(A) and cos(B). Since sin(A) = 1/4 and A is in the second quadrant (QII), we can apply the Pythagorean identity sin^2(A) + cos^2(A) = 1 to find cos(A). Rearranging the equation gives:
cos^2(A) = 1 - sin^2(A)
cos^2(A) = 1 - (1/4)^2
cos^2(A) = 1 - 1/16
cos^2(A) = 15/16
Taking the square root of both sides, we get:
cos(A) = sqrt(15)/4
Similarly, since sin(B) = -1/2 and B is in the fourth quadrant (QIV), we can find cos(B) using the same steps:
cos^2(B) = 1 - sin^2(B)
cos^2(B) = 1 - (-1/2)^2
cos^2(B) = 1 - 1/4
cos^2(B) = 3/4
cos(B) = sqrt(3)/2
Now, we can use the identity for tan(A+B):
tan(A+B) = (tan(A) + tan(B)) / (1 - tan(A) * tan(B))
We already have sin(A), cos(A), sin(B), and cos(B). To find tan(A) and tan(B), we can use the definitions of tangent:
tan(A) = sin(A) / cos(A) = (1/4) / (sqrt(15)/4) = 1 / sqrt(15)
tan(B) = sin(B) / cos(B) = (-1/2) / (sqrt(3)/2) = -1 / sqrt(3)
Plugging these values into the formula for tan(A+B):
tan(A+B) = (1 / sqrt(15) + (-1 / sqrt(3))) / (1 - (1 / sqrt(15)) * (-1 / sqrt(3)))
Simplifying further:
tan(A+B) = (sqrt(3) - sqrt(15)) / (sqrt(45) - sqrt(45))
tan(A+B) = (sqrt(3) - sqrt(15)) / 0
tan(A+B) = Undefined
Therefore, tan(A+B) is undefined.