Question 818283
"compute the exact values of sin(2θ), cos(2θ), tan(2θ) without a calculator.
<pre>
 cos(&#952;)={{{-4/5}}}, {{{pi/2}}} < &#952; < {{{pi}}}.
's
I changed your x's to &#952; so that x could represent the values on the
x-axis rather than the angle.

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We draw the angle &#952; in the second quadrant.


{{{drawing(200,200,-6,6,-6,6,line(-10,0,10,0),line(0,-10,0,10),
green(line(0,0,-4,3)),red(arc(0,0,3,-3,0,144)),locate(.2,2.3,theta) )}}}

And we draw a perpendicular to the x-axis from the end of the green line
forming a right triangle.

{{{drawing(200,200,-6,6,-6,6,line(-10,0,10,0),line(0,-10,0,10),
green(line(0,0,-4,3),line(-4,3,-4,0)),red(arc(0,0,3,-3,0,144)),locate(.2,2.3,theta), locate(-2.4,0,x), locate(-4.6,2,y),
locate(-2.2,2.8,r)


 )}}}
The bottom side of the triangle is x
The vertical side of the triangle is y, 
and the green line is r.

The cosine is {{{x/r}}} and we are given that  cos(&#952;)={{{-4/5}}}
so we will take x as the numerator, considering it negative, since it
goes to the left.  So we will label it x=-4, and we will label r as
the denominator of {{{-4/5}}}, considering it positive, since it goes
up.  So we will label it as r=5

{{{drawing(200,200,-6,6,-6,6,line(-10,0,10,0),line(0,-10,0,10),
green(line(0,0,-4,3),line(-4,3,-4,0)),red(arc(0,0,3,-3,0,144)),locate(.2,2.3,theta), locate(-3.4,0,x=-4), locate(-4.6,2,y),
locate(-2.6,2.8,r=5) )}}}

We calculate y from

r² = x²+y²
5² = (-4)²+y²
25 = 16+y²
 9 = y²
±3 = y, we take the positive value since y goes up.

So we label the y-value as 3

{{{drawing(200,200,-6,6,-6,6,line(-10,0,10,0),line(0,-10,0,10),
green(line(0,0,-4,3),line(-4,3,-4,0)),red(arc(0,0,3,-3,0,144)),locate(.2,2.3,theta), locate(-3.4,0,x=-4), locate(-5.9,2,y=3),locate(-2.6,2.8,r=5) )}}}

sin(2&#952;) = 2sin(&#952; )cos(&#952; ) 

Since sin(&#952; ) = {{{y/r}}}, sin(&#952; ) = {{{3/5}}}  

sin(2&#952;) = 2sin(&#952; )cos(&#952; ) = 2{{{(3/5)(-4/5)}}} = {{{-24/25}}}

cos(2&#952;) = cos²(&#952;)-sin²(&#952;) = {{{(-4/5)^2}}}-{{{(3/5)^2}}} = {{{16/25}}}-{{{9/25)}}} = {{{7/25}}} 

tan(2&#952;) = {{{(2tan(theta))/(1-tan^2(theta))}}} 

Since the tangent is {{{y/x}}}, tan(&#952;) = {{{3/(-4)}}} = {{{-3/4}}}

tan(2&#952;) = {{{(2(-3/4))/(1-(-3/4)^2)}}} = {{{(-3/2)/(1-9/16)}}} 

Multiply top and bottom by 16

tan(2&#952;) = {{{(16*expr(-3/2))/(16(1-9/16))}}} = {{{(-24)/(16-9) }}} = {{{(-24)/7}}} = {{{-24/7}}}

Edwin</pre>