Trigonometry help

glenn101 · 2008-12-20 18:38:25

Hi everyone, I'm stuck with these trig questions, I don't understand why some give more solutions than others over the required domain it just doesn't make sense to me.
I also need help with solving tan equations, I've never done them before I've just began next years specialist work early but the book doesn't explain it.

1. Solve each of the following for x element [0,2pi];
sinx=-root3/2
sintheta=pi/3
theta=pi/3, (2pi-pi/3), (pi+pi/3)
theta= pi/3, 4pi/3, 5pi/3
in the answer it is just
4pi/3, 5pi/3 why is that?? why isn't pi/3 a solution?
and
2cos(2x)=-1 0≤x≤2pi (not sure if this part is correct my maths teacher taught me it.)
cos(2x)= -1/2 0≤2x≤4pi
costheta=pi/3
theta=pi/3, (pi-pi/3), (pi+pi/3)
= 2pi/3, 4pi/3
the answer in the book is
pi/3, 2pi/3, 4pi/3, 5pi/3 this is really confusing me. Why do they now have the pi/3 and where do they get the 5pi/3 from?

2. Solve each of the following for x element[0,2pi];
a) 2tan(x/2)+2=0
2tan(x/2)=-2
tan(x/2)=-1
from there I just lose it.
Thanks in advance for any help,
Glenn.

Last edited by glenn101 (2008-12-21 14:19:33)

Jai Ganesh · 2008-12-20 19:22:32

For (2), x=3pi/4 or 7pi/4. For these values of x, tanx= -1.

For (1), if sinx = -√(3/2), x=4pi/3, 5pi/3

If 2 cos(2x)=-1, cos(2x)=-1/2, 2x=2pi/3, 4pi/3

glenn101 · 2008-12-21 14:12:02

thanks for the fast reply ganish, much appreciated.
however, for question 1b)
If 2cos(2x)=-1 the answer is pi/3, 2pi/3, 4pi/3, 5pi/3 why is that?
and how are your deriving your answers in two steps, if there is a technique I'd really like to know, it would help alot.

Jai Ganesh · 2008-12-22 01:41:32

glenn101,
In early stages of learning trignometry, it would very useful to remember the values of sin, cos, tan, sec, cosec and cot for 0°, 30°, 45°, 60°, 90°, 180° and 360°.

2 cos(2x)=-1 implies cos(2x) = -1/2. Therefore, if 2x=2pi/3, 4pi/3, x=pi/3, 2pi/3.

I hope I am not thrusting too much upon you but also try to have the graphs of sinx, cox, tanx for 0° to 360°, if possible, 0° to 720°.

glenn101 · 2008-12-22 12:33:19

Ah, you mean exact values?
we only know 0°, 30°, 45°, 60°, 90° we havn't been taught anything above, thats next year.
Thanks for the ongoing help ganesh, its much appreciated.
Actually Ganesh I see what I've been doing wrong I understand 100% now, thanks for putting up with my confusion.

Last edited by glenn101 (2008-12-22 14:00:17)

Jai Ganesh · 2008-12-22 16:06:46

glenn101, I don't mean the exact values, the numbers like

etc. for sin 0, 30, 45, 60, 90 etc.

mathsyperson · 2008-12-23 05:11:22

An interesting pattern happens if you rewrite those numbers like this:

You might find them easier to remember that way.

glenn101 · 2008-12-23 13:55:32

That is an interesting pattern indeed!, will definatly help, requires minimal memory!
thanks for sharing that mathsy!

Macy · 2008-12-24 13:22:32

To answer your question 1(b), why that?

2cos(2x) = -1 0<= 2x <= 4pi
cos (2x) = -1/2
2x = 2pi/3 and 4pi/3 (you need to look at unit circle to figure out which quadrants giving you negative cos from 0 to 2pi, in this case is 2nd and 3rd)

because the domain restriction from 0<= 2x <= 4pi therefore you will need to add another 2pi to obtain all possible solutions to satisfy domain restriction so the final value will be

2x = 2pi/3, 4pi/3, 2pi/3 +2pi, 4pi/3 + 2pi

x = pi/3, 2pi/3, 4pi/3, 5pi/3

okie enjoy Christmas break, do not work so hard okie

Math Is Fun Forum

#1 2008-12-20 18:38:25

Trigonometry help

#2 2008-12-20 19:22:32

Re: Trigonometry help

#3 2008-12-21 14:12:02

Re: Trigonometry help

#4 2008-12-22 01:41:32

Re: Trigonometry help

#5 2008-12-22 12:33:19

Re: Trigonometry help

#6 2008-12-22 16:06:46

Re: Trigonometry help

#7 2008-12-23 05:11:22

Re: Trigonometry help

#8 2008-12-23 13:55:32

Re: Trigonometry help

#9 2008-12-24 13:22:32

Re: Trigonometry help

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