Question 198049
Find the value of the five other trig functions, given 

{{{sinX=-sqrt(2)/2}}} and {{{cosX<0}}} also 

give degree and radian measure of X
<pre><font size = 4 color = "indigo"><b>
I will use a capital X to represent the angle, and a 
small x to represent the x-coordinate in the graph.
Don't get them confused!

Since
{{{sin(X)=(-sqrt(2))/2}}}

and that is a negative number, this tells us that 
the angle X is in either QIII of QIV.

Since
{{{cosx<0}}}

that tells us that the cosine is a negative number,
and we know that the cosine is negative in QII and QIII,

so therefore we know that angle X is in QIII, so we
draw it:

{{{drawing(400,400,-3,3,-3,3,
graph(400,400,-3,3,-3,3), triangle(-sqrt(2),-sqrt(2),-sqrt(2),0,0,0),
locate(-.7,.3,x), locate(-1.6,-.6,y),locate(-.7,-.7,r),
rectangle(-sqrt(2)+.2,-.2,-sqrt(2),0),blue(arc(0,0,.8,-.8,0,225) ),
locate(-.2,.6,X)

)}}}

Since the sine is {{{y/r}}}, we take the
numerator of the given sine, which is {{{-sqrt(2)}}} to be the 
value of y, and the denominator 2 to be the value of r.

{{{drawing(400,400,-3,3,-3,3,
graph(400,400,-3,3,-3,3), triangle(-sqrt(2),-sqrt(2),-sqrt(2),0,0,0),
locate(-.7,.3,x), locate(-2,-.6,y=sqrt(2)),locate(-.6,-.6,r=2),
rectangle(-sqrt(2)+.2,-.2,-sqrt(2),0),blue(arc(0,0,.8,-.8,0,225) ),
locate(-.2,.6,X)


)}}}

Next we calculate the value of x (small x) using the Pythagorean
theorem:

{{{x^2+y^2=r^2}}}
{{{x^2+(-sqrt(2))^2=(2)^2}}}
{{{x^2+2=4}}}
{{{x^2=2}}}
{{{x=""+-sqrt(2)}}}

We know to take the velue of x as negative, because
x goes left from the origin, so
{{{x=-sqrt(2)}}}

and we label it thusly:

{{{drawing(400,400,-3,3,-3,3,
graph(400,400,-3,3,-3,3), triangle(-sqrt(2),-sqrt(2),-sqrt(2),0,0,0),
locate(-1.1,.3,x=-sqrt(2)), locate(-2,-.6,y=sqrt(2)),locate(-.6,-.6,r=2),
rectangle(-sqrt(2)+.2,-.2,-sqrt(2),0),blue(arc(0,0,.8,-.8,0,225) ),
locate(-.2,.6,X) )}}}

Now we get the other 5 trigonometric functions by using
the definitions:

{{{cos(X)=x/r=(-sqrt(2))/2=-sqrt(2)/2}}}

{{{sec(X)=r/x=2/(-sqrt(2))=-2/sqrt(2)=(-2sqrt(2))/(sqrt(2)sqrt(2))=(-2sqrt(2))/2=-sqrt(2)}}}

{{{csc(X)=r/y=2/(-sqrt(2))=-2/sqrt(2)=(-2sqrt(2))/(sqrt(2)sqrt(2))=(-2sqrt(2))/2=-sqrt(2)}}}

{{{tan(X)=y/x=(-sqrt(2))/(-sqrt(2))=1}}}

{{{cot(X)=r/x=(-sqrt(2))/(-sqrt(2))=1}}}

Since the right triangle has two legs equal it is isosceles
are the reference angle is therefore 45°.

Since the angle is in QIII, we must add 180° to its
reference angle, so

the angle X in degree measure is 45°+180° or 225°

In radian measure, we multiply by {{{pi/180^o}}}

{{{(225^o)(pi/180^o)=5pi/4 }}}


Edwin</pre>