Alright so ive been talking to my math tutor from the univ. about this single question for about a week now and i HAVE to get it done.
If and
For a second quadrant angle and a third quadrant angle
Have to find
Alright so ive been talking to my math tutor from the univ. about this single question for about a week now and i HAVE to get it done.
If and
For a second quadrant angle and a third quadrant angle
Have to find
Construct one triangle that refers to and one that refers to .
Since is in quadrant II, we let the side opposite the angle have length 7 and the side adjacent to the angle have length -24
So that means that the hypotenuse of our triangle will have length
Therefore, we can conclude that and .
Does this make sense so far?
You should do something similar for .
Then once you have a right triangle, it is easy to find , and the like.
If this doesn't make sense, post back. To help it sink in, try to draw the triangle I described to you above.
My suggestion would be at first stage forget the sign, just find the trig values. e.g., tan A = 7/24 we know that the hypotenuse will be 25. so we will get sin A = 7/25; cosA = 23/25, tan A = 7/24; cot A = 24/7, sec A = 24/25 and csec A = 25/7.
Now in second step see where that angle lies, in this case A lies in II quadrant, we know that in second quadrant sine and its reciprocal csec are positive and all rest are negative. Put signs accordingly and you have the result:
sin A = 7/25; cosA = - 23/25, tan A = - 7/24; cot A = - 24/7, sec A = - 24/25 and csec A = 25/7.