Easy enough but apparently my math tutor doesnt understand that I DONT GET IT.

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October 27th 2009, 08:24 AM
Djaevel

Easy 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 $Tan\alpha = {- 7\over24}$ and $Cot\beta = {3\over4}$

For a second quadrant angle $\alpha$ and a third quadrant angle $\beta$

Have to find

$Sin(\alpha + \beta) , Cos (\alpha + \beta) , Tan (\alpha + \beta), Sin(\alpha - \beta) , Cos(\alpha - \beta) , Tan(\alpha - \beta)$
October 27th 2009, 08:39 AM
Chris L T521

Quote:

Originally Posted by Djaevel

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 $Tan\alpha = {- 7\over24}$ and $Cot\beta = {3\over4}$

For a second quadrant angle $\alpha$ and a third quadrant angle $\beta$

Have to find

$Sin(\alpha + \beta) , Cos (\alpha + \beta) , Tan (\alpha + \beta), Sin(\alpha - \beta) , Cos(\alpha - \beta) , Tan(\alpha - \beta)$

Construct one triangle that refers to $\alpha$ and one that refers to $\beta$ .

Since $\alpha$ 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 $\sqrt{(-24)^2+7^2}=25$

Therefore, we can conclude that $\sin \alpha =\frac{7}{25}$ and $\cos\alpha = \frac{-24}{25}$ .

Does this make sense so far?

You should do something similar for $\beta$ .

Then once you have a right triangle, it is easy to find $\sin\beta$ , $\cos\beta$ 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, 08:55 AM
Djaevel

Ohhh i was like trying to use.. well nevermind lol dont know what i was doing/thinking >< thanks haha
October 27th 2009, 09:03 AM
Djaevel

So for the $\beta$ triangle it would be like x=-3 and y=-4 ? then get hyp and go from there?
October 27th 2009, 09:06 AM
Chris L T521

Quote:

Originally Posted by Djaevel

So for the $\beta$ triangle it would be like x=-3 and y=-4 ? then get hyp and go from there?

(Yes)
March 3rd 2013, 06:16 PM
ThomasHornbeck

Re: Easy enough but apparently my math tutor doesnt understand that I DONT GET IT.

try Search: math - Tutor Universe
March 3rd 2013, 07:55 PM
ibdutt

Re: 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.