Evaluate the expression under the given conditions.
sin (A+B); tan A= 12/5, A in quadrant 3, sin B= -3sqrt(10)/10, B in quadrant 4.
To evaluate the expression sin(A + B) given the conditions tan A = 12/5 (A in quadrant 3) and sin B = -3√10/10 (B in quadrant 4), we need to find the values of sin A and cos A first.
Since tan A = 12/5, we can determine the opposite and adjacent sides of the right triangle in quadrant 3 using the Pythagorean theorem.
Let's assume the opposite side is 12x and the adjacent side is 5x.
Using the Pythagorean theorem, we have:
(5x)^2 + (12x)^2 = hypotenuse^2
25x^2 + 144x^2 = hypotenuse^2
169x^2 = hypotenuse^2
√(169x^2) = hypotenuse
13x = hypotenuse
So, the hypotenuse is 13x. Since hypotenuse ≠ 0, we can divide the opposite side by the hypotenuse to find sin A:
sin A = (opposite side / hypotenuse) = (12x / 13x) = 12/13.
Next, we need to determine the sign of sin A in quadrant 3. Since A is in quadrant 3, the sine function is negative. Therefore, sin A = -12/13.
Moving on to sin B, which is given as -3√10/10. Since sin B is negative and B is in quadrant 4, we need to find the absolute value of sin B, which is 3√10/10.
Now with the values of sin A and sin B, we can evaluate sin(A + B) using the trigonometric identity:
sin(A + B) = sin A * cos B + cos A * sin B
But since we don't have the value of cos A available, we need to find it.
Using the Pythagorean identity, we have:
cos^2 A = 1 - sin^2 A
cos^2 A = 1 - (-12/13)^2
cos^2 A = 1 - 144/169
cos^2 A = (169 - 144)/169
cos^2 A = 25/169
Taking the square root of both sides, we have:
cos A = ± √(25/169)
cos A = ± 5/13
Since A is in quadrant 3, where cos A is negative, we have:
cos A = -5/13
Now we can calculate sin(A + B):
sin(A + B) = sin A * cos B + cos A * sin B
sin(A + B) = (-12/13) * (-3√10/10) + (-5/13) * (3√10/10)
sin(A + B) = (36√10/130) + (-15√10/130)
sin(A + B) = (36√10 - 15√10) / 130
sin(A + B) = 21√10/130
Therefore, sin(A + B) is equal to (21√10/130).
in QIII, sin and cos are both negative.
You have a 5-12-13 triangle, so
sinA = -12/13
cosA = -5/13
in QIV, cos is positive.
You have a 3-1-√10 triangle, so
cosB = 1/√10
Now apply the addition formula
sin(A+B) = sinAcosB + cosAsinB