Question 1206406
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I'll replace x with the Greek letter theta (symbol {{{theta}}}) since it is more commonly used with trig angles. 
Also it's to avoid potential upcoming confusion when I'll use x in a different way.


Given: {{{tan(theta) = -4/3}}} where angle {{{theta}}} is in quadrant II.


Goal: Compute {{{sin(2theta)}}}, {{{cos(2theta)}}}, and {{{tan(2theta)}}}


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Quadrant II is the northwest quadrant i.e. upper left corner.
This is for any point (x,y) such that x < 0 and y > 0.



Here's what the diagram looks like
{{{
drawing(400,400,-6,6,-6,6,
graph(400,400,-6,6,-6,6),
line(0,0,-3,0),
line(-3,0,-3,4),
line(-3,4,0,0),
locate(-4,3,"y"),
locate(-5.5,2,"opposite"),
locate(-2,-1+0.3,"x"),
locate(-2.7,-1.5+0.3,"adjacent"),
locate(-2+0.7,3,"r"),
locate(-2.7+0.7,4,"hypotenuse"),
locate(0.6,1,theta),
arc(0,0,1.5,1.5,233.1301023542,360),
locate(-5.5,5.5,matrix(1,2,"Quadrant","II")),
locate(-5.5,-5,matrix(1,4,"Diagram","is","to","scale"))
)
}}}


Tangent is the ratio of opposite over adjacent. 
{{{"tan(angle)" = "opposite"/"adjacent"}}}
This is the same as saying {{{tan(theta) = y/x}}}
y = opposite
x = adjacent


If {{{tan(theta) = -4/3}}} then we would have y = 4 and x = -3 so the x < 0 and y > 0 requirements are fulfilled.
The terminal point is located at (x,y) = (-3,4).


Now let's find the hypotenuse of this right triangle using the Pythagorean theorem. 
{{{a^2+b^2 = c^2}}}
{{{x^2+y^2 = r^2}}}
{{{r = sqrt(x^2+y^2)}}}
{{{r = sqrt((-3)^2+(4)^2)}}}
{{{r = sqrt(25)}}}
{{{r = 5}}}
The hypotenuse is 5 units long.
We have a 3-4-5 right triangle.


y = opposite = 4
x = adjacent = -3
r = hypotenuse = 5


which means,
{{{sin(theta) = y/r}}}
{{{sin(theta) = 4/5}}}
and
{{{cos(theta) = x/r}}}
{{{cos(theta) = -3/5}}}



Furthermore,
{{{sin(2theta) = 2*sin(theta)*cos(theta)}}}
{{{sin(2theta) = 2*(4/5)*(-3/5)}}}
{{{sin(2theta) = -24/25}}} <font color=red>which is one of the answers</font>
and also
{{{cos(2theta) = cos^2(theta) - sin^2(theta)}}} see <font color=green>note</font> below
{{{cos(2theta) = (-3/5)^2 - (4/5)^2}}}
{{{cos(2theta) = 9/25 - 16/25}}}
{{{cos(2theta) = -7/25}}} <font color=red>which is another answer</font>
lastly
{{{tan(2theta) = (sin(2theta))/(cos(2theta))}}} see <font color=green>note</font> below
{{{tan(2theta) = matrix(1,3,sin(2theta),"divide",cos(2theta))}}}
{{{tan(2theta) = matrix(1,3,(-24/25),"divide",(-7/25))}}}
{{{tan(2theta) = (-24/25)*(-25/7)}}}
{{{tan(2theta) = 24/7}}} <font color=red>which is the last answer we need</font>


<font color=green>Note</font>: There are other identities you could use as an alternative pathway.
Refer to this list of trig identities
<a href="https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf">https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf</a>
Specifically refer to the "double angle formulas" section on page 2.


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A summary of the answers:
{{{sin(2theta) = -24/25}}}
{{{cos(2theta) = -7/25}}}
{{{tan(2theta) = 24/7}}}
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