Question 1205032
If sin(x)=-5/13 and x is in quadrant III,...
<pre>
So for angle x, we have a 5-12-13 right triangle in quadrant III,
where <italics>x</italics>=-12, y=-5, and r=+13.

There is often a conflict of notation when x is used for an angle, and also
for values of the adjacent side of the defining right triangle. There is a
problem here. But I think you won't get confused.  Teachers aren't always
careful to point out this conflict, which happens a lot. (just like the problem
of how to say "the sign of the sine". <font face="wingdings" size=4>J</font> )

So cos(x)=x/r=(-12)/(+13)=-12/13 and tan(x)=y/x=(-5)/(-12)=+5/12

{{{180^o<x<270^o}}}
{{{90^o<x/2<135^o}}} so x/2 is in quadrant II (the upper half of quadrant II).

So sin(x/2) is positive, cos(x/2) is negative, and tan(x/2) is negative.

So we just use the half-angle formulas:

{{{sin(x/2)="" +- sqrt((1-cos(x))/2)}}}, {{{cos(x/2)="" +- sqrt((1+cos(x))/2)}}}

We know to use the + because x/2 is in quadrant II, where sine is positive.

{{{sin(x/2)= sqrt((1-(-12/13))/2)=sqrt((1+12/13)/2)= sqrt((13+12)/26)=sqrt(25/26)=5/sqrt(26)=5sqrt(26)/26}}}

We know to use the - because x/2 is in quadrant II, where cosine is negative.

{{{cos(x/2)=-sqrt((1+(-12/13))/2)=-sqrt((1-12/13)/2)=-sqrt((13-12)/26)=-sqrt(1/26)=-1/sqrt(26)=-sqrt(26)/26}}}

Now we could use a formula for tan(x/2), but now all we need is {{{tan(x/2)=sin(x/2)/cos(x/2)}}}

{{{tan(x/2)=sin(x/2)/cos(x/2)= (5sqrt(26)/26)/(-sqrt(26)/26)=(5cross(sqrt(26)/26))/(-cross(sqrt(26)/26))=-5}}}

Edwin</pre>