Question 1201902
first of all, you don't have parentheses where they belong.
that makes it difficult to see what the values really are.
i rewrote the values using parentheses to make it clearer.


after rewriting, i get the expression to be solved is:
sin(arccos(-sqrt(7)/5)
and i get the possible solutions are:
4 * sqrt(2) / 5
-4 * sqrt(2) / 5
3 * sqrt(2) / 5
-3 * sqrt(2) / 5


let A be the angle.
arccos(-sqrt(7)/5) means that cos(A) = (-sqrt(7)/5).
since cosine is equal to adjacent side divided by hypotenuse, you get:
adjacent side is -sqrt(7) and hppotenuse is 5.
the adjacent side is the x value on the graph.
the opposite side is the y value on the graph.
you know that the adjacent side is negative, so the angle will be either in the second quadrant or the third quadrant.
the hypotenuse is always positive.
you can use pythaorus to find the value of the opposite side.
pythgorus says tha adjacent side squared plus opposite side squared = hypotenuse squared.
you get:
x^2 + y^2 = h^2.
x represents the adjacent side.
y represens the opposite side.
h represents the hypotenuse of thetriangle formed.
A represents the angle.
since you know the value of x and the value of h, the formula becomes:
(-sqrt(7))^2 + y^2 = 5^2
simplify to get:
7 + y^2 = 25
solve for y^2 to get y^2 = 18
solve for y to get y = plus or minus sqrt(18).
if y is positive, then the angle is in the second quadrant.
if y is negative, then the angle is in the third quadrant
here'smy diagram of how i see the graphical representation.
note that angle A in the diagram is the reference angle.
in this problem, you really don't need to solve for the reference angle.
the problem can be solvedby the other values you are given.


A is your reference angle.
in the second quadrant, your angle would be 180 minus A.
in the third quadrant, your angle would be 180 plus A.


your original expression was sin(arccos(-sqrt(7)/5)
we said that arccos(-sqrt(7)/5) means that cos(A) = -sqrt(7)/5.
from that information, and from drawing the diagram, we were able to find that the adjacent side of angle A was -sqrt(7) and the hypotenuse was 5.
from that information, we were able to determine that the opposite side was either sqrt(18) or - sqrt(18).
if the angle was in the second quadrant, then opposite side was sqrt(18).
if the angle was in the third quadrant, then opposite side ws -sqrt(18).


if we simplify sqrt(18), we find that sqrt(18) = 3 * sqrt(2).
therefore sin(A) was equal to 3 * sqrt(2) if the angle was in the second quadrant and sn(A) was equal to -3 * sqrt(2) if the angle was in the third quadrant.


i beieve your solution is that sin(arccos(-sqrt(7)/5)) is either 3 * sqrt(2) / 5 or -3 * sqrt(2) / 5.


here's my sketch.


<img src = "http://theo.x10hosting.com/2023/042601.jpg">


here's my graph of cos(x) which shows the angles in the second quadrant (121...   degrees) and the third quadrant (238... degrees) have the same cosine value.


<img src = "http://theo.x10hosting.com/2023/042602.jpg">


here's my graph of sin(x) which shows the angles in the second quadrant (121... degrees) and the third wuadrant (238... degrees) have the same sine value, but with different signs (positive in the second quadrant and negative in the third quadrant).


<img src = "http://theo.x10hosting.com/2023/042603.jpg">


if all this seems confusing, don't feel bad, because it is confusing and very difficult to figure out unless you've had experience in working out problems such as these.


i'm reasonably confident the answers i gave you are correct.
let me know if you have any questions.
theo