sin(arctan(-4/3)

okay so i made arctan(-4/3) = x
so im solving for sin(x)
and x= arctan(-4/3)
so then tanx = -4/3

and now im stuck i don't know what to do after that

Draw the right triangle (3,4,5)

3^2+4^2 = 9+16 = 25 = 5^2
angle A = atan(4/3)
sin A = 4/5
now look for quadrant
tan = -4/3 in quadrant 2 and in quadrant 4
in quadrant 2 sin A = 4/5
in quadrant 4 sin A = -4/5

just draw the triangle

tan = y/x
sin = y/h

h = 5 (3-4-5 triangle!)

principal value of arctan(-4/3) is in QIV.

sin = -4/5

To find sin(arctan(-4/3)), you first need to determine the value of tan(x) = -4/3 and then use knowledge of trigonometric functions to calculate sin(x).

Since tan(x) = -4/3, we can set up a right triangle where the opposite side is -4 and the adjacent side is 3.

Now, use the Pythagorean theorem to find the hypotenuse of the triangle:
hypotenuse^2 = opposite^2 + adjacent^2
hypotenuse^2 = (-4)^2 + 3^2
hypotenuse^2 = 16 + 9
hypotenuse^2 = 25
hypotenuse = √25
hypotenuse = 5

Now that we know the lengths of the sides of the triangle, we can find the value of sin(x). Remember that sin(x) is the ratio of the opposite side to the hypotenuse.

sin(x) = opposite/hypotenuse
sin(x) = -4/5

Therefore, sin(arctan(-4/3)) = -4/5.