Question 969905
How do I find sin and cos of theta if tangent of theta equals -1?
THANK YOU!!
<pre>
The tangent is negative in the 2nd and 4th quadrants.  Its referent angle
is the first quadrant angle that has +1 for its tangent.  That's {{{"45°"}}} or {{{pi/4}}}.  
Its sine is {{{sqrt(2)/2}}} and its cosine is also {{{sqrt(2)/2}}}

So to get the smallest 2nd quadrant positive solution, we either

subtract 45° from 180° to get it into the 2nd quadrant, and get 
180°-45° = 135°.  Its sine is {{{sqrt(2)/2}}} and it's cosine is {{{-sqrt(2)/2}}}.

or we

subtract {{{pi/4}}} from {{{pi}}} to get it into the 2nd quadrant, 
and get 
{{{pi-pi/4 = 3pi/4}}}, and its sine is {{{sqrt(2)/2}}} and its cosine is {{{-sqrt(2)/2}}}

That's the 2nd quadrant solutions worked out in both degrees and radians.

------

To get the smallest 4th quadrant positive solution, we either

subtract 45° from 360° to get it into the 4th quadrant, and get 
360°-45° = 315°,

or we

subtract {{{pi/4}}} from {{{2pi}}} to get it into the 4th quadrant, 
and get {{{2pi-pi/4 = 8pi/4-pi/4 = 7pi/4}}}


{{{pi-pi/4 = 3pi/4}}}.

The sine is {{{""-sqrt(2)/2}}}, and the cosine is {{{""+sqrt(2)/2}}}

That's the 4th quadrant solutions worked out in both degrees and radians.  

So if theta is in the 2nd quadrant its sine is {{{""+sqrt(2)/2}}} and its 
cosine is {{{-sqrt(2)/2}}}, and

if theta is in the 4th quadrant its sine is {{{-sqrt(2)/2}}} and its cosine 
is {{{""+sqrt(2)/2}}}.

Edwin</pre>