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

- October 27th 2009, 09:24 AMDjaevelEasy enough but apparently my math tutor doesnt understand that I DONT GET IT.
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

- October 27th 2009, 09:39 AMChris L T521
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. - October 27th 2009, 09:55 AMDjaevel
Ohhh i was like trying to use.. well nevermind lol dont know what i was doing/thinking >< thanks haha

- October 27th 2009, 10:03 AMDjaevel
So for the triangle it would be like x=-3 and y=-4 ? then get hyp and go from there?

- October 27th 2009, 10:06 AMChris L T521
- March 3rd 2013, 07:16 PMThomasHornbeckRe: Easy enough but apparently my math tutor doesnt understand that I DONT GET IT.
- March 3rd 2013, 08:55 PMibduttRe: Easy enough but apparently my math tutor doesnt understand that I DONT GET IT.
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.