Trigonometric Equations!!! (1 Viewer)

[adonis1]

0≤ Θ ≤ 360

Tan^2 3Θ=1 (^2 means to the power of 2 i.e squared)

please.... for the love of god...someone help...and please be descriptive..unlike my math teacher..

 

[FinalFantasy]

0≤ Θ ≤ 360

Tan² 3Θ=1
tan 3Θ=+-1
tan pi\4 is 1 so just do the angles
3Θ=pi\4, 3pi\4,5pi\4,7pi\4,9pi\4, 11pi\4..........
therefore Θ=pi\12, pi\4, 5pi\12, 7pi\12, 3pi\4, 11pi\12............
do it until Θ gets to 360 or 2pi

 

[adonis1]

lol..um wtf is pi. do u mean like pi as in 3.14...etc Cause if u do im lost, if it isnt can u tell me wat it is

 

[FinalFantasy]

oh sorry lol, just remember pi=180 degrees in dis case

 

[adonis1]

ty, im still lost but dont worry i got maths 2morow and ill ask my teacher, i appreciate ur help, u must be pretty smart lol sorry i couldnt get it

 

[FinalFantasy]

u know the four quadrants?
if the angle is in the first quadrant is where all sin\cos\tan are positive
e.g tan 45, cos 45, sin 45, dey are all in the first quadrant and are positive.
2nd quadrant is where only sin is positive, e.g sin 135=1\root 2 but cos 135=-1\root 2
3rd one is where tan is positive
4th one is where cos is positive

in ur q. u have tan 3Θ=+-1, so the tan can be positive or negative...
u know tan of an angle is 1, that angle must be 45 right?
so 3Θ=45, den u go around the four quadrants getting all the 45 degrees measured from the x-axis
then to get Θ u just divide by three...

and im not "smart" to be able to do dis, it is from practise, u need to be able to get these easily

 

[RyddeckerSMP]

The key thing for this question is to change the domain for 0< theta <360, to 0< 3 theta < 1080....... then you can answer the question as you normally would, but you have to remember that you do not stop at 360, like you normally would as the domain for the equation has been changed........

 

[Trev]

adonis1; I will do it in terms of degree angles for you:
(using part of FinalFantasy's work)
0≤ Θ ≤ 360

Tan² 3Θ=1
tan 3Θ=+-1
Therefore, the basic angle lies in all 4 quadrants (because where tan is both positive and negative - from 'CAST' or 'All Stations To Central' on the unit circle)
basic angle is 45 degrees

0≤ 3Θ ≤ 1080
Therefore....
3Θ = 45°, 135°, 225°, 315°, 405°, 495°, 585°, 675°, 765°, 855°, 945°, 1035°
Θ = 15°, 45°, 75°, 105°, 135°, 165°, 195° , 225°, 255°, 285°, 315°, 345°

 

[Trev]

The key thing for this question is to change the domain for 0< theta <360, to 0< 2 theta < 1080....... .

Correction; change the domain to 3Θ, not 2Θ - just a typo, but making sure the guy doesn't get confused.

 

[adonis1]

adonis1; I will do it in terms of degree angles for you:
(using part of FinalFantasy's work)
0≤ Θ ≤ 360

Tan² 3Θ=1
tan 3Θ=+-1
Therefore, the basic angle lies in all 4 quadrants (because where tan is both positive and negative - from 'CAST' or 'All Stations To Central' on the unit circle)
basic angle is 45 degrees

0≤ 3Θ ≤ 1080
Therefore....
3Θ = 45°, 135°, 225°, 315°, 405°, 495°, 585°, 675°, 765°, 855°, 945°, 1035°
Θ = 15°, 45°, 75°, 105°, 135°, 165°, 195° , 225°, 255°, 285°, 315°, 345°

na i get it now, after i asked my math teacher she explained it, i understood all the quadrants before that anyway, what i didnt understand was why there were 12 answers not 6, she said it was 12 because it was +- which ment all the 4 quadrants multiplied by 3. I just got lost on the +- part but i get it. Cheers everyone for the help anyway

 

[Trev]

Never forget the ± as I often did.