• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Solving trig. functions (1 Viewer)

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
How do I do what it says in the yellow box? How do I get 90 degrees, 270 degrees, etc.

 

calamebe

Active Member
Joined
Mar 19, 2015
Messages
462
Gender
Male
HSC
2017
Well for this question, cos((pi/4)t)=0. When cos(x) is 0, x is pi/2, 3pi/2, 5pi/2 etc. So when solving cos((pi/4)t)=0, we need (pi/4)t to be equal to pi/2, 3pi/2, 5pi/2, and 7pi/2, as I'm guessing the domain was restricted. So then we solve for t to get t=2, 6, and 10.
 

keepLooking

Active Member
Joined
Aug 25, 2014
Messages
477
Gender
Male
HSC
2015
Many (including me) were taught by the phrase "All Stations To Central" or ASTC.





 

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
Many (including me) were taught by the phrase "All Stations To Central" or ASTC.


I know how to get 90, and 270 by using ASTC, but I don't know how to get further than that.
 

calamebe

Active Member
Joined
Mar 19, 2015
Messages
462
Gender
Male
HSC
2017
I know how to get 90, and 270 by using ASTC, but I don't know how to get further than that.
So is it that you don't know how to get angles greater than 360? If so, what you do is instead of starting at 0 degrees, as you normally do, you can start at 360 degrees. Or you could get the basic angles you need, in this case 90 and 270, and then add 360 degrees to them, to get 450 and 630. If the domain is greater, for instance 0<x<1080, then you would also add 720 to the angles.
 

keepLooking

Active Member
Joined
Aug 25, 2014
Messages
477
Gender
Male
HSC
2015
I know how to get 90, and 270 by using ASTC, but I don't know how to get further than that.
It is two revolutions, basically when you know cos theta = 0, theta can be 90, 270, 450 ....,
so you just add 360 for a full revolution to start at the first quadrant again.
 

BlueGas

Well-Known Member
Joined
Sep 20, 2014
Messages
2,448
Gender
Male
HSC
N/A
It is two revolutions, basically when you know cos theta = 0, theta can be 90, 270, 450 ....,
so you just add 360 for a full revolution to start at the first quadrant again.
So is it that you don't know how to get angles greater than 360? If so, what you do is instead of starting at 0 degrees, as you normally do, you can start at 360 degrees. Or you could get the basic angles you need, in this case 90 and 270, and then add 360 degrees to them, to get 450 and 630. If the domain is greater, for instance 0<x<1080, then you would also add 720 to the angles.
How about if sinx = 1/2? x = 30, x = 150, and what do I do now?
 

calamebe

Active Member
Joined
Mar 19, 2015
Messages
462
Gender
Male
HSC
2017
How about if sinx = 1/2? x = 30, x = 150, and what do I do now?
Are you looking for angles greater than 360? Well add 360 to them, to get x = 390 and x = 510. Also have you learnt general solutions. I don't know whether they are 2 unit or 3 unit but they are useful in calculating angles greater than 360 and less than 0.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top