If tan A= 3/4 and cos B=-5/13, where A and B are both third-quadrant angles, find sin(A+B).

Please explain how you found all the answers

make sketches of two right angles triangles in the third quadrants

since tanA = 3/4, the terminal arm of angle A ends at (-4,-3) , and r = 5
so sinA = -3/5, and cosA = -4/5
(you have a 3-4-5 right-angled triangle)

since cosB = -5/13 , the terminal arm of angle B ends at (-5,-12), and r = 13
so sinB = -12/13
(you have the 5-12-13 right-angled triangle)

we know
sin(A+B) = sinAcosB + cosAsinB
= (-3/5)(-5/13) + (-4/5)(-12/13)
= 15/65 + 48/65
= 63/65

To find sin(A+B), we can use the trigonometric identity: sin(A+B) = sin(A)cos(B) + cos(A)sin(B).

Given that A is in the third quadrant and tan(A) = 3/4, we can determine the values of sin(A) and cos(A) as follows:

We know that tan(A) = sin(A) / cos(A), so we have:
3/4 = sin(A) / cos(A)

Multiplying both sides by cos(A), we get:
3cos(A) = 4sin(A)

Using the Pythagorean identity sin^2(A) + cos^2(A) = 1, we can apply it to the equation above:
3cos(A) = 4sin(A) = 3cos^2(A) + 4sin^2(A)

Since A is in the third quadrant, both sin(A) and cos(A) will be negative. Let's assume that cos(A) = -x and sin(A) = -y (where x and y are positive values), then:

3(-x) = 4(-y)
-3x = -4y
x = (4/3)y

Using the identity sin^2(A) + cos^2(A) = 1, we can substitute x and y:
(4/3)y^2 + y^2 = 1
(16/9)y^2 + (9/9)y^2 = 1
(25/9)y^2 = 1
y^2 = 9/25
y = ±√(9/25) = ±3/5

Since A is in the third quadrant, sin(A) is negative. Therefore, sin(A) = -3/5, and cos(A) = -4/5.

Similarly, if B is in the third quadrant and cos(B) = -5/13, we can determine the values of sin(B):
Using the Pythagorean identity sin^2(B) + cos^2(B) = 1:
sin^2(B) + (-5/13)^2 = 1
sin^2(B) + 25/169 = 1
sin^2(B) = 1 - 25/169 = 144/169
sin(B) = ±√(144/169) = ±12/13

Since B is in the third quadrant, sin(B) is negative. Therefore, sin(B) = -12/13.

Now, we have sin(A) = -3/5, cos(A) = -4/5, sin(B) = -12/13, and cos(B) = -5/13.

Using the trigonometric identity sin(A+B) = sin(A)cos(B) + cos(A)sin(B), we can substitute the values to find sin(A+B):
sin(A+B) = (-3/5)(-5/13) + (-4/5)(-12/13)

simplifying, we get:
sin(A+B) = 15/65 + 48/65
sin(A+B) = 63/65

Therefore, sin(A+B) equals 63/65.

To find sin(A+B), we can use the trigonometric identity sin(A+B) = sin(A)cos(B) + cos(A)sin(B).

Let's find sin A and cos A first. In the third quadrant, the sine and cosine values are negative. We are given that tan A = 3/4, which means that sin A = (3/4) / √(1 + (3/4)^2) and cos A = 1 / √(1 + (3/4)^2).

sin A = (3/4) / √(1 + 9/16) = (3/4) / √(25/16) = (3/4) / (5/4) = 3/5
cos A = 1 / √(1 + 9/16) = 1 / √(25/16) = 1 / (5/4) = 4/5

Next, let's find sin B and cos B. We are given that cos B = -5/13, which means that sin B = √(1 - (cos B)^2). Since B is also a third-quadrant angle, sin B is negative.

sin B = -√(1 - (-5/13)^2) = -√(1 - 25/169) = -√(144/169) = -12/13

Now, we can substitute these values into the formula sin(A+B) = sin(A)cos(B) + cos(A)sin(B).

sin(A+B) = (3/5)(-5/13) + (4/5)(-12/13)
= -15/65 - 48/65
= -63/65

Therefore, sin(A+B) is -63/65.