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trig help!


beli16

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is this SL? I don't wanna go doing HL stuff and get it wrong :) trig wasn't my specialty.. if it's SL I can probably do it...

I'm trying to do it now.. dunno if I can..

hey does the question say it wants exact figures? if not use ur graphing software and just do it on ur calc :)

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maybe u have it set as degrees instead of radians... (your calc) as for the mathematical thing.. first convert the degree value u have there into radians, and that would be the first one for the other two.. erm you have to figre out which quartile the angle is on (I think it's the 2nd quartile) and since this is tan then you'll have to subtract what you get with (pi). I don't really remember the quartile stuff really well since we could use GDC for paper one, if u have it in ur notes just review them (I kinda threw out all my math notes :) )

EDIT wait no! lol add (pi) to ur answer, don't subtract it, they're 1st and 3rd quartile :) not 2nd :)

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  • 2 months later...
Simultaneous equation!

My mind's not sufficiently with me at the moment, but as with intersecting 'lines' on the GDC graph, you can set up a simultaneous equation and solve.

hmmm i dnt see how this can be solved using simultaneous equation... if you could please type it, wud be nice to know different methods.

now beli16, you said that tan-1x=78.69

bt i believe you ment t tan-1(5)=78.69 since you take tan^-1 for both sides to have that x=78.69, now that is your critical angle. We know that tan should be positive since 5 is positive, thus your angle must be either in the first quadrant or the third quadrant. If we are using degrees, then your answer would be: x = 78.69 and x = 258.69

now if we are using radians then we go back to beginning at tan-1(5) in radians that would be = 1.37 thus making your critical angle 1.37. Again tan is positive thus the angle lies in 1st and 3rd quadrants so for the first quadrant x = 1.37, for the 3rd quadrant x = pi+1.37 = 4.51

Now it is a matter of what your limits are, you say that you should solve for 0<x<10. Here i get kinda unsure of whether i solved wrng or whether the limits are incorrect since if the limits are correct then the question should be solved in radians, if solved in radians then you have two answers since both are above 0 and lower than 10. As for our degrees answers they wont apply and thats why you shoudl use radians. plz recheck the limits and let me know what the answr is...

my trig is kinda rusty we covered it first semester last year... better get back and revise :D

hope i helped :P

Edited by HMSChocolate
Please, no text speak...
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have a question which i just cant figure outt!!!

sinx=5cosx

i can get that tanx=5

and that tan-1x=78.69

now i have to solve on 0<x<10...how do i do that??

i know the answers are: 1.37, 4.515, 7.66...

thxxx :P

this question is rather easy. first, set your calculator mode to radian and find arctan(5) and your answer will be 1.37.

because tan is positive in the first and third quadrant, we need to find the value of tan in the third quadrant. the value is : 1.37 + pi = 4.515

now, the limit for x is 0<x<10. that means the value may exceed 2pi. so, we can get the third value of x which is 1.37 + 2pi = 7.66.

note that we do not take 4.515 + 2pi = 10.798 as a value of x because it is out of the limit.

in the end, the values of x are 1.37, 4.515, 7.66

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this question is rather easy. first, set your calculator mode to radian and find arctan(5) and your answer will be 1.37.

because tan is positive in the first and third quadrant, we need to find the value of tan in the third quadrant. the value is : 1.37 + pi = 4.515

now, the limit for x is 0<x<10. that means the value may exceed 2pi. so, we can get the third value of x which is 1.37 + 2pi = 7.66.

note that we do not take 4.515 + 2pi = 10.798 as a value of x because it is out of the limit.

in the end, the values of x are 1.37, 4.515, 7.66

i dnt get it??? why did u add 1.37 to 2pi... simply on the basis that x can exceed 2 pi?

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i dnt get it??? why did u add 1.37 to 2pi... simply on the basis that x can exceed 2 pi?

Well, 2pi is 360 degree or one revolution. So, the value of x can be more than 360 degree/2pi. When I add 1.37 to 2pi means that in a new revolution there will be another value of x such that tanx=5. The "angle between the x-axis and line in the unit circle" (sorry I forgot the right term) is the same, 1.37. To get the angle, we add 2pi to 1.37.

sorry i can't explain well cause it's hard to tell math without showing things (diagram, work) in front of you. Besides, when explaining math, my english went bad.

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