Angle a lies in quadrant 2, and tan(a)= -(12/5). Angle B lies in quadrant 4, and cos(B)= (3/5).
What is the exact value of sin(a+B) ?
tanA = -12/5 = y/x, so in quad II, y = 12, x = -5
r^2 = x^2 + y^2 = 25+144 = 169
r = 13
sinA = 12/13, cosA = -5/13
cos(B)= (3/5) = x/r , where B is in quad IV, so x=3, r=5
x^2 + y^2 = r^2
9 + y^2 = 25
y = -4 ,
sinB = -4/5, cosB = 3/5
sin(A+B) = sinAcosB + cosAsinB
= (12/13)(3/5) + (-5/13)(-4/5)
= .....
To find the exact value of sin(a+B), we need to use the trigonometric identities and the given information.
First, we can use the information about angle a and tangent to find the values of sin(a) and cos(a). Since angle a lies in quadrant 2 and tangent is negative, we know that cosine is positive in that quadrant. Therefore, cos(a) = √(1 - sin^2(a)) = √(1 - ((12/5)^2)) = √(1 - (144/25)) = √(25/25 - 144/25) = √(-119/25).
Next, we can use the information about angle B and cosine to find the value of sin(B). Since angle B lies in quadrant 4 and cosine is positive, we know that sine is negative in that quadrant. Therefore, sin(B) = -√(1 - cos^2(B)) = -√(1 - ((3/5)^2)) = -√(1 - (9/25)) = -√(25/25 - 9/25) = -√(16/25) = -√(4/5).
Now, we can use the sum-of-angle identity for sine to find sin(a+B):
sin(a+B) = sin(a)cos(B) + cos(a)sin(B).
Plugging in the values we found:
sin(a+B) = (√(-119/25))((3/5)) + (√(4/5))(-√(4/5)).
Simplifying further:
sin(a+B) = (-√(357)/25) + (-4/5).
Combining like terms:
sin(a+B) = (-√(357) - 4)/5.
Therefore, the exact value of sin(a+B) is (-√(357) - 4)/5.