Author Topic: Circular functions help? (1/2 Methods) (Read 4184 times) Tweet Share

Andiio · « **on:** August 28, 2010, 09:45:52 pm »

Teacher has not gone over the method to which you solve trigonometric equations very well... could someone please explain the steps in which to solve trig equations?

E.g.

1. Solve each of the following for x 'is an element of' [0, 2pi]:

(a) sinx = 1

Thanks in advance!

pooshwaltzer · « **Reply #1 on:** August 28, 2010, 09:49:04 pm »

TT/2 and its southerly cousin 3TT/2

Andiio · « **Reply #2 on:** August 28, 2010, 09:50:17 pm »

Quote from: pooshwaltzer on August 28, 2010, 09:49:04 pm

TT/2 and its southerly cousin 3TT/2

Could you please explain the method in which to solve these types of questions?

TyErd · « **Reply #3 on:** August 28, 2010, 09:53:01 pm »

two T's for a pi, why did i never think of that

pooshwaltzer · « **Reply #4 on:** August 28, 2010, 09:55:35 pm »

sinx = 1
x = sin-1(1)
x = 90 (TT/2)
Keep adding TT...
next is x = 90 + 180 = 270 (3TT/2)
then comes x = 90 + 2(180) = 450 (5TT/2) ... BUT wait, that's past our domain of [0, 2pi] so TT/2 and 3TT/2 it is.

Andiio · « **Reply #5 on:** August 28, 2010, 10:00:47 pm »

Quote from: pooshwaltzer on August 28, 2010, 09:55:35 pm

sinx = 1
x = sin-1(1)
x = 90 (TT/2)
Keep adding TT...
next is x = 90 + 180 = 270 (3TT/2)
then comes x = 90 + 2(180) = 450 (5TT/2) ... BUT wait, that's past our domain of [0, 2pi] so TT/2 and 3TT/2 it is.

When you add the pi, is it always adding the period of the given domain?

98.40_for_sure · « **Reply #6 on:** August 28, 2010, 10:04:04 pm »

Quote from: pooshwaltzer on August 28, 2010, 09:55:35 pm

sinx = 1
x = sin-1(1)
x = 90 (TT/2)
Keep adding TT...
next is x = 90 + 180 = 270 (3TT/2)
then comes x = 90 + 2(180) = 450 (5TT/2) ... BUT wait, that's past our domain of [0, 2pi] so TT/2 and 3TT/2 it is.

sin(3TT/2) = -1

There is only one solution in [0,2TT]

brightsky · « **Reply #7 on:** August 28, 2010, 10:04:50 pm »

Quote from: Andiio on August 28, 2010, 09:45:52 pm

Teacher has not gone over the method to which you solve trigonometric equations very well... could someone please explain the steps in which to solve trig equations?

E.g.

1. Solve each of the following for x 'is an element of' [0, 2pi]:

(a) sinx = 1

Thanks in advance!

$\sin(x) = 1$

Using exact values and symmetry properties on the unit circle, we get:
$x = ..., -\frac{7\pi}{2}, -\frac{3\pi}{2}, \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, ...$

But we have $x \in [0, 2\pi]$ , and so the only solutions within that domain are:

$x = \frac{\pi}{2}$

Note: $\sin\left(\frac{3\pi}{2}\right) = - 1$ [Beaten xD]

Andiio · « **Reply #8 on:** August 28, 2010, 10:06:43 pm »

Quote from: brightsky on August 28, 2010, 10:04:50 pm

Quote from: Andiio on August 28, 2010, 09:45:52 pm
Teacher has not gone over the method to which you solve trigonometric equations very well... could someone please explain the steps in which to solve trig equations?

E.g.

1. Solve each of the following for x 'is an element of' [0, 2pi]:

(a) sinx = 1

Thanks in advance!

$\sin(x) = 1$

Using exact values and symmetry properties on the unit circle, we get:
$x = ..., -\frac{7\pi}{2}, -\frac{3\pi}{2}, \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, ...$

But we have $x \in [0, 2\pi]$ , and so the only solutions within that domain are:

$x = \frac{\pi}{2}$

Note: $\sin\left(\frac{3\pi}{2}\right) = - 1$ [Beaten xD]

How do you work out the 4pi difference between the x values? :/

pooshwaltzer · « **Reply #9 on:** August 28, 2010, 10:08:12 pm »

Quote from: brightsky on August 28, 2010, 10:04:50 pm

[Beaten xD]

Not at all. Your lucid and succinct LaTeX rendition is much more coherent than my effigy of a firing squad line-up.

98.40_for_sure · « **Reply #10 on:** August 28, 2010, 10:08:38 pm »

The 4pi difference?... in one revolution of the unit circle, there is one solution. Thus to find recurring solutions, you simply add/minus 2pi (the period)

Andiio · « **Reply #11 on:** August 28, 2010, 10:12:36 pm »

Quote from: 99.95_for_sure on August 28, 2010, 10:08:38 pm

The 4pi difference?... in one revolution of the unit circle, there is one solution. Thus to find recurring solutions, you simply add/minus 2pi (the period)

OH oops my bad, I looked at it wrong xD

How would you work something like:

Solve each of the following for x 'is an element of' [-TT, 2TT]
sinx = 1

and

Solve each of the following for x 'is an element of' [-180 degrees,360 degrees]:
sin x = -1/2

98.40_for_sure · « **Reply #12 on:** August 28, 2010, 10:17:00 pm »

1) the answer would be the same

2) [-180d,360d] = [-TT,2TT]
Sin is negative in 3rd and 4th quadrants.
Thus solutions are: -5TT/6, -TT/6, 7TT/6, 11TT/6

brightsky · « **Reply #13 on:** August 28, 2010, 10:18:38 pm »

There is only really two things you need to understand in order to solve trig equations: exact values and symmetry properties (plus taking note of the domain). So for example, if we have:

Find all solutions of $\sin(\pi + x) = \frac{\sqrt{2}}{2}$ given $x \in [-\pi, \pi]$ .

First take heed of the domain.

$-\pi \le x \le \pi$

$\therefore 0 \le x + \pi \le 2\pi$ (Add $\pi$ to all sides.) ----> This step is VERY IMPORTANT when you have something other than "x" inside the trig.

Then treat the thing in the "sine" (in this case $\pi + x$ as one single entity). If you want..let $a = x + \pi$ .

So $a \in [0,2\pi]$ .

Now $\sin(a) = \frac{\sqrt{2}}{2}$ ...and by our exact values, we know that $a$ must be $\frac{\pi}{4}$ , or some "symmetrical angle" to it.

By our unit circle, we know that $\sin(x) = \sin(\pi - x) = \sin(2\pi + x) = \sin(2\pi + \pi - x) = ....$ (These can be negative as well but given our domain in this case, we are not interested in the negatives.)

That means that $a = \frac{\pi}{4}, \pi - \frac{\pi}{4} , 2\pi + \frac{\pi}{4}, ....$

But we want only the solutions in the domain of a, which is in this case $[0, 2\pi]$ .

So the only solutions for $a$ is $\frac{\pi}{4}, \pi - \frac{\pi}{4}$ .

Simplifying that, we have $a = \frac{\pi}{4}, \frac{3\pi}{4}$

But we need to find $x$ . We know that $a = \pi + x$ , so then $x = a - \pi$ .

Hence $x = \frac{\pi}{4} - \pi, \frac{3\pi}{4} - \pi$

$x = -\frac{3\pi}{4}, -\frac{\pi}{4}$ .

|ll|lll| · « **Reply #14 on:** August 28, 2010, 10:22:28 pm »

$sin(x) = \frac{-1}{2}$

$Reference angle = 30 deg$
Since sin(x) is negative, solutions are in third and fourth quadrant.
$x = 180+30, 360-30$
$x = 210, 330$
Now, we have to take into account the negative domain [-180,0)
$x = -30, 0-(180-30)$
$x= -30, -150$
Hence, x = -150, -30, 210, 330 degrees

Search the forums now!

We have moved!

Author Topic: Circular functions help? (1/2 Methods) (Read 4184 times) Tweet Share

Andiio

Circular functions help? (1/2 Methods)

pooshwaltzer

Re: Circular functions help? (1/2 Methods)

Andiio

Re: Circular functions help? (1/2 Methods)

TyErd

Re: Circular functions help? (1/2 Methods)

pooshwaltzer

Re: Circular functions help? (1/2 Methods)

Andiio

Re: Circular functions help? (1/2 Methods)

98.40_for_sure

Re: Circular functions help? (1/2 Methods)

brightsky

Re: Circular functions help? (1/2 Methods)

Andiio

Re: Circular functions help? (1/2 Methods)

pooshwaltzer

Re: Circular functions help? (1/2 Methods)

98.40_for_sure

Re: Circular functions help? (1/2 Methods)

Andiio

Re: Circular functions help? (1/2 Methods)

98.40_for_sure

Re: Circular functions help? (1/2 Methods)

brightsky

Re: Circular functions help? (1/2 Methods)

|ll|lll|

Re: Circular functions help? (1/2 Methods)

Recent Posts

User:
Password: