Question 390489
<pre>
If tan2x = - 24/7, what are the values of sin x and cos x?

To avoid conflict of notation I'm going to replace your x by <font face = "symbol">q</font>


{{{Tan(2theta) = - 24/7}}}

{{{(2Tan(theta))/(1-Tan^2theta) = (-24)/7}}}

Cross multiply:

{{{14Tan(theta)=-24(1-Tan^2theta)}}}

Distribute:

{{{14Tan(theta)=-24+24Tan^2theta}}}

Get 0 on the right

{{{-24Tan^2theta + 14Tan(theta) + 24 = 0}}}

Divide both sides through by -2

{{{12Tan^theta - 7Tan(theta) - 12 = 0}}}

Factor

{{{(3Tan(theta)-4)(4Tan(theta)+3)=0}}}

Use 0 factor property on the first factor:

{{{3Tan(theta)-4=0}}}

{{{3Tan(theta) = 4}}}

{{{Tan(theta) = 4/3}}}

This could either be in the 1st quadrant or the 3rd since
the tangent is positive in the 1st and 3rd quadrants.

So angle {{{theta}}} could be either of these:

{{{drawing(200,200,-3,7,-5,5, graph(200,200,-3,7,-5,5), line(0,0,3,4),
line(3,4,3,0), line(0,0,3,0), locate(3.2,2.5,y=4), locate(1,1,x=3) ) }}}{{{drawing(200,200,-7,3,-5,5, graph(200,200,-7,3,-5,5), line(0,0,-3,-4),
line(-3,-4,-3,0), line(0,0,-3,0), locate(-5.3,-1.5,y=-4), locate(-3,1,x=-3) ) }}}

In either case r = 5

{{{r^2=x^2+y^2}}}
{{{r^2=3^2+4^2}}}
{{{r^2=9+16}}}
{{{r^2=25}}}
{{{r=5}}}

{{{r^2=x^2+y^2}}}
{{{r^2=(-3)^2+(-4)^2}}}
{{{r^2=9+16}}}
{{{r^2=25}}}
{{{r=5}}}

So we have

{{{drawing(200,200,-3,7,-5,5, graph(200,200,-3,7,-5,5), line(0,0,3,4),
line(3,4,3,0), line(0,0,3,0), locate(3.2,2.5,y=4), locate(1,1,x=3),locate(.2,3,r=5) ) }}}{{{drawing(200,200,-7,3,-5,5, graph(200,200,-7,3,-5,5), line(0,0,-3,-4),
line(-3,-4,-3,0), line(0,0,-3,0), locate(-5.3,-1.5,y=-4), locate(-3,1,x=-3),locate(-3,-4,r=5)  ) }}}

For the first graph {{{sin(theta) = y/r=4/5}}}, {{{cos(theta)=x/r=3/5}}}
For the second graph {{{sin(theta) = y/r=-4/5}}}, {{{cos(theta)=x/r=-3/5}}}


Use 0 factor property on the second factor:

{{{4Tan(theta)+3=0}}}

{{{4Tan(theta) = -3}}}

{{{Tan(theta) = -3/4}}}

This could either be in the 2nd quadrant or the 4th since
the tangent is negative in the 2nd and 4th quadrants.

So angle {{{theta}}} could be either of these:

{{{drawing(200,200,-7,3,-5,5, graph(200,200,-7,3,-5,5), line(0,0,-4,3),
line(-4,3,-4,0), line(0,0,-4,0), locate(-5.6,2.3,y=3), locate(-3.6,1,x=-4) ) }}}{{{drawing(200,200,-3,7,-5,5, graph(200,200,-3,7,-5,5), line(0,0,4,-3),
line(4,-3,4,0), line(0,0,4,0), locate(4.1,-1.6,y=-3), locate(1,1,x=4) ) }}}

In either case r = 5

{{{r^2=x^2+y^2}}}
{{{r^2(-4)^2+3^2}}}
{{{r^2=16+9}}}
{{{r^2=25}}}
{{{r=5}}}

{{{r^2=x^2+y^2}}}
{{{r^2=4^2+(-3)^2}}}
{{{r^2=16+9}}}
{{{r^2=25}}}
{{{r=5}}}

So we have

{{{drawing(200,200,-7,3,-5,5, graph(200,200,-7,3,-5,5), line(0,0,-4,3),
line(-4,3,-4,0), line(0,0,-4,0), locate(-5.6,2.3,y=3), locate(-3.6,1,x=-4),

locate(-3,3,r=5)


 ) }}}{{{drawing(200,200,-3,7,-5,5, graph(200,200,-3,7,-5,5), line(0,0,4,-3),
line(4,-3,4,0), line(0,0,4,0), locate(4.1,-1.6,y=-3), locate(1,1,x=4), 
locate(1,-2.3,r=5)

) }}}

For the first graph {{{sin(theta) = y/r=3/5}}}, {{{cos(theta)=x/r=-4/5}}}
For the second graph {{{sin(theta) = y/r=-3/5}}}, {{{cos(theta)=x/r=4/5}}}

Edwin</pre>