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Expert: - 2/27/2009


Question
Hi my question is how would i solve this
sin a=40/41,-3pie/2 ,<a<-pie; tanB= square root-6, pie/2 <B<pie

Answer
Note that pie is really pi.  That's a greek letter.  
Pie is something that tastes yummy.  Had any lately?

Note that a calculator will give the answer in the first quadrant.
Note that the answer should be in quadrant II - 2pi.
The 2pi is subtracted of to make it negative.

To find the number in quadrant II, find y=pi-x, where x is the answer is quadrant 1.  To put it where they want the answer, then subtract 2pi from that value of y.


Is that tanB = square root(-6)?  Negatives don't have real squareroots.   It looks like its really just a dash to designate
that is what your taking the squareroot of, though.  Or maybe it is suppose to be -squareroot(6) since the answer is in quadrant II.
In this quadrant, the tan() values are negative.
To find the answer, take the arctan(-root(6)) and subtract it from pi.

You need to know what the trig functions look like.
Here are the values at 0, pi/2, pi, 3pi/2, and 2pi.

sin(): 0 1 0 -1 0
cos(): 1 0 -1 0 1
tan(): 0 -- 0 -- 0

Note that the -- means the tangent of that value is undefined.
On the left side, it reaches forever upward.  On the right side, it comes from ever downward.  This makes tan() and always increasing function whereas sin() and cos() go up and down smoothly.

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