Question 142454
Use the given information to find {{{sin(2phi)}}}, {{{cos(2phi)}}}, {{{tan(2phi)}}}. 
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Formulas needed are

{{{sin(2phi)= 2sin(phi)cos(phi)}}} 
{{{cos(2phi)=2cos^2(phi)-1}}}
{{{tan(2phi)=sin(2phi)/cos(2phi) }}}

1. {{{tan(phi)=1/2}}}, {{{pi < phi < (3pi)/2)}}} 

That's the THIRD quadrant.

Draw a picture of angle theta in the third quadrant

Since {{{TANGENT = (OPPOSITE)/(ADJACENT) = y/x}}} 

If theta were in the first quadrant, we'd let the 
numerator of {{{1/2}}} be y and the denominator of 
{{{1/2}}} be x.  But since theta is in the third 
quadrant, we have to make x and y both negative.  
So take y, the opposite, as the negative of the 
numerator, -1, and take x to be the negative of 
the denominator, -2. 

So plot the point (-2,-1) in the third quadrant and 
draw a radius to it and draw a perpendicular up to 
the x-axis, forming a right triangle:

{{{drawing(300,300,-3,3,-3,3,
graph(300,300,-3,3,-3,3), triangle(0,0,-2,-1,-2,0),
locate(-1.5,.5,"x=-2"), locate(-2.7,-.4,"y=-1")
 )}}}

Next we need to find r, the hypotenuse. We use the
Pythagorean theorem:

{{{r^2=x^2+y^2}}}

{{{r^2=(-2)^2+(-1)^2}}}

{{{r^2=4+1}}}

{{{r^2=5}}}

{{{r=sqrt(5)}}}

So we label that:

{{{drawing(300,300,-3,3,-3,3,
graph(300,300,-3,3,-3,3), triangle(0,0,-2,-1,-2,0),
locate(-1.5,.5,"x=-2"), locate(-2.7,-.4,"y=-1"), locate(-1.2,-.5,r=sqrt(5))
 )}}}

Now we have {{{x=-2}}},{{{y=-1}}},{{{r=sqrt(5)}}}
And so 
{{{sin(phi)=y/r=(-1)/(sqrt(5))=-sqrt(5)/5}}}
{{{cos(phi)=y/r=(-2)/(sqrt(5))=-2sqrt(5)/5}}}

Substituting in

{{{sin(2phi)= 2sin(phi)cos(phi)}}} 
{{{sin(2phi)= 2(-sqrt(5)/5)(-2sqrt(5)/5)}}}
{{{sin(2phi)= 2(2(5))/25 = 20/25=4/5  }}}

Substituting in

{{{cos(2phi)=2cos^2(phi)-1}}}
{{{cos(2phi)=2((-2sqrt(5))/5)^2-1}}}
{{{cos(2phi)=2( 4*5 )/25 )-1}}}
{{{cos(2phi)=40/25-1}}}
{{{cos(2phi)=8/5-1}}}
{{{cos(2phi)=8/5-5/5}}}
{{{cos(2phi)=3/5}}}

Substituting those values in:
{{{tan(2phi)=sin(2phi)/cos(2phi) }}}
{{{tan(2phi)=(4/5)/(3/5) }}}
{{{tan(2phi)=(4/5)(5/3)}}}
{{{tan(2phi)=(4/cross(5))(cross(5)/3)}}}
{{{tan(2phi)=4/3}}}

If I have time I'll come back and do the others.

Edwin</pre>