Question 28402: Please help me solve this equation=>
-780 (degrees) in six triginometric ratio
sin, cos tan, cot, sec, csc
I put it on a graph and i know it goes counterclock wise since it is a (-) after this i landed in the 4th quadrant then i -360+ -360+ -60=-780(degrees)
but i don't know where to go from there or if all this work i have done is right at all. If someone could please try and help me out or guide me in the right direction I would much be appreciative
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! Please help me solve this equation=>
-780 (degrees) in six triginometric ratio
sin, cos tan, cot, sec, csc
We know that any trignometic fn of [2n(Pi)+ or minus(angle theta)]
= + or minus the angle theta
Pi radians = 360 degrees
Therefore to (-780)degrees add the nearest positive number of revolutions namely two revoultions in this problem that is add 2X(360)=720 degrees
Then (-780) + 720 = (-60)degrees.
What is the idea?
The idea is to bring the angle to an acute form (or in some cases obtuse form)
numerically.
As you have rightly understood- the position of the angle (-780) is the fourth quadrant and we are in a situation where
any trig fn(-780) is equivalent to the same trig fn (-60)
Therefore sin (-780 degrees) =sin(-60) -sin(60) = -(sqrt3)/2
cos(-60)= +cos(60) = 1/2
tan(-60)= -tan(60) = -(sqrt3)
cosec(-60)= -cosec(60) = -(1/sin(60)) = -1/[(sqrt3)/2] = -2/(sqrt3)
sec(-60)= +sec(60) = +(1/cos(60)) = 1/(1/2) = 2
cot(-60)= -cot(60) = -1/(tan(60))= -1/(sqrt3)
Note: In the fourth quadrant only the cosine and its reciprocal fn secant are positive.
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