Minimum Value

mathfun · Dec 29, 2005

huh?!?!? i'm so confused!!!

:?

is this right?

2(sin(x/2)cos(pi/3) - sin(pi/3)cos(x/2)
=2sin((x/2) - (pi/3))

(x/2) - (pi/3) =

x = (11pi)/6 ??

like that? i understand y we use (3pi)/2...but i don't understnad... y not use -2?
and i never new that for.... sin(alpha - beta) where (alpha - beta) = minimum??

Unco · Dec 29, 2005

mathfun said:
so then it should be:

2(sin(x/2)cos(pi/3) - sin(pi/3)cos(x/2)
=2sin((x/2) - (pi/3))

(x/2) - (pi/3) =

x = (11pi)/6 ??

like that? i understand y we use (3pi)/2...but i don't understnad... y not use -2?
and i never new that for.... sin(alpha - beta) where (alpha - beta) = minimum??

You have established

. . . \(\displaystyle \L y = 2\sin{(\frac{x}{2} - \frac{\pi}{3})}\)

is minimum at \(\displaystyle y = -2\).

So set \(\displaystyle y = -2\) and solve for \(\displaystyle x\):

. . . \(\displaystyle \L 2\sin{(\frac{x}{2} \, - \, \frac{\pi}{3})} = -2\)

. . . \(\displaystyle \L \sin{(\frac{x}{2} \, - \, \frac{\pi}{3})} = -1\)

Some practise (ie. a long and ugly way):

Sine is negative in quadrants 3 and 4:

Code:

         pi/2 
  s       |         a
          |
          |
pi -------+-------- 0, 2pi
       | /|\ |
       \/ | \/
       /  |  \      c  
  t       |
         3pi/2

And \(\displaystyle \L \arcsin{(1)} = \frac{\pi}{2}\)

So, solving for \(\displaystyle \L \frac{x}{2} - \frac{\pi}{3}\) first:

. . . \(\displaystyle \L \begin{align*} \left(\frac{x}{2} \, - \, \frac{\pi}{3}\right) &= \pi \, + \, \frac{\pi}{2} \\
&= \frac{3\pi}{2}\\
\end{align*}\)

. . . (and \(\displaystyle \L \left(\frac{x}{2} - \frac{\pi}{3}\right) = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}\))

Now solve for \(\displaystyle x\):

. . . \(\displaystyle \L \left(\frac{x}{2} \, - \, \frac{\pi}{3}\right) = \frac{3\pi}{2}\)

. . . \(\displaystyle \L \frac{x}{2} = \frac{3\pi}{2} + \frac{\pi}{3}\)

. . . \(\displaystyle \L \frac{x}{2} = \frac{11\pi}{6}\)

. . . \(\displaystyle \L x = \frac{11\pi}{3}\)

If there were two solutions for \(\displaystyle \left(\frac{x}{2} \, - \, \frac{\pi}{3}\right)\), we solve for \(\displaystyle x\) as above for each solution.

mathfun · Dec 29, 2005

why is it not
x/2 - pi/3 = pi/2

why do you have to add pi to pi/2?
and what is 2pi - pi/2? (the one in brakets)

in addition, i don't understnad your last sentence
thanx for the help, unco.

tkhunny · Dec 29, 2005

mathfun said:
what are you talking about? what is your point here?
i was just asking how you knew what to do to know to use that identity, etc. if u don't know how u knew yourself, u can just say that. its okay.

It is very difficult to answer your question directly. Through years of experience, and the practice or observation of many methodologies, I just saw it. There is a really helpful self-deprecating phrase in the medical community. Their training phrase is something like this, "If you don't think of it, you won't diagnose it." They are trying to diagnose illness and we are trying to solve problems. It is the same kind of thought process. If you see it, it will be obvious how to proceed. If you don't see it, you will have no idea how to proceed. This is one of the great blockades to solving problems - finding some way to get started. It instills fear in the hearts of many beginning students. I have used several things over the years.

1) WRITE DOWN clear and concise definitions.

That doesn't help us on this problem.

1) WRITE DOWN every thing you know about _______ (fill in the blank).

This may help on this problem. If you write down various relationships between sine and cosine, you may happen to write down the formula for the sine of a sum. That would be great IF you see it.

1) Just try ANYTHING. You can't break it. It's just an equation or expression.

This is often very helpful. You do NOT have to see the entire solution before you start doing SOMETHING. You may learn something almost by accident, but really you can blame your courage for whatever you happen to learn.

Good luck. It's not always obvious how to proceed.

mathfun · Dec 29, 2005

tkhunny said:
mathfun said:

what are you talking about? what is your point here?
i was just asking how you knew what to do to know to use that identity, etc. if u don't know how u knew yourself, u can just say that. its okay.

Click to expand...

It is very difficult to answer your question directly. Through years of experience, and the practice or observation of many methodologies, I just saw it. There is a really helpful self-deprecating phrase in the medical community. Their training phrase is something like this, "If you don't think of it, you won't diagnose it." They are trying to diagnose illness and we are trying to solve problems. It is the same kind of thought process. If you see it, it will be obvious how to proceed. If you don't see it, you will have no idea how to proceed. This is one of the great blockades to solving problems - finding some way to get started. It instills fear in the hearts of many beginning students. I have used several things over the years.

1) WRITE DOWN clear and concise definitions.

That doesn't help us on this problem.

1) WRITE DOWN every thing you know about _______ (fill in the blank).

This may help n this problem. If you write down various relationships between sine and cosine, you may happen to write down the formula for the sine of a sum. That would be great IF you see it.

1) Just try ANYTHING. You can't break it. It's just an equation or expression.

This is often very helpful. You do NOT have to see the entire solution before you start doing SOMETHING. You may learn something almost by accident, but really you can blame your courage for whatever you happen to learn.

Good luck. It's not always obvious how to proceed.

thanks so much for the tips, tkhunny. i will keep them in mind. thanks.

mathfun · Dec 29, 2005

mathfun said:
why is it not
x/2 - pi/3 = pi/2

why do you have to add pi to pi/2?
and what is 2pi - pi/2? (the one in brakets)

is it because it has to be in quadrants III and IV?
if yes, then i understna dthe pi+pi/2

but why 2pi-pi/2?

Unco · Dec 29, 2005

mathfun said:
why do you have to add pi to pi/2?
and what is 2pi - pi/2? (the one in brakets)

Look at the c.a.s.t diagram. When we are in quadrant 3, we must add pi to the principal angle* (arcsin(1)); and when in quadrant 4, we subtract the principle angle from 2pi.

mathfun said:
in addition, i don't understand your last sentence

Say we had \(\displaystyle \L \sin{\left(\frac{x}{2} \, + \, \frac{\pi}{6}\right) = -\frac{1}{2}\)

We solve for \(\displaystyle \left(\frac{x}{2} \, + \, \frac{\pi}{6}\right)\) first:

-1/2 is negative so we are in the 't' and 'c' quadrants:

Code:

         pi/2 
  s       |         a
          |
          |
pi -------+-------- 0, 2pi
       | /|\ |
       \/ | \/
       /  |  \      c  
  t       |
         3pi/2

We calculate the principal angle by taking the arcsin of positive 1/2:
. . . \(\displaystyle \L \arcsin{(\frac{1}{2})} = \frac{\pi}{6}\)

The two solutions are in quadrants 3 ('t') and 4 ('c') as on the diagram.

The solution in quadrant 3 is found by adding pi to our principal angle:
. . . \(\displaystyle \L \left(\frac{x}{2} \, + \, \frac{\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\)

The solution in quadrant 4 is found by subtracting our principal angle from 2pi:
. . . \(\displaystyle \L \left(\frac{x}{2} \, + \, \frac{\pi}{6}\right) = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}\)

So our solutions are
. . . \(\displaystyle \L \left(\frac{x}{2} \, + \, \frac{\pi}{6}\right) = \frac{7\pi}{6}, \, \frac{11\pi}{6}\)

We now solve for \(\displaystyle x\) for each separately:

. . . \(\displaystyle \L \left(\frac{x}{2} \, + \, \frac{\pi}{6}\right) = \frac{7\pi}{6}\)

. . . \(\displaystyle \L \frac{x}{2} = \frac{7\pi}{6} -
\frac{\pi}{6} = \pi\)

. . . \(\displaystyle \L x = \frac{5\pi}{3}\)
or

. . . \(\displaystyle \L \left(\frac{x}{2} \, + \, \frac{\pi}{6}\right) = \frac{11\pi}{6}\)

. . . \(\displaystyle \L \frac{x}{2} = \frac{11\pi}{6} - \frac{\pi}{6} = \frac{5\pi}{3}\)

. . . \(\displaystyle \L x = \frac{10\pi}{3}\)

* I don't think this is correct terminology, but . . .

mathfun · Dec 29, 2005

thanx a lot unco!!
so then there are two answers to my question as well??

let me summarize:
if i want it to be in QUAD III, i add pi.
if i want it to be in QUAD IV, i minus it by 2pi.

what about QUAD I and II?
i'm guessing
QUADI...i add 2 pi??
QUAD II..i minus pi??

Unco · Dec 29, 2005

There's just the one solution in the domain 0 to 2pi for the original problem. We expect that because there is only one minimum point of a sine curve in this domain.

That's shown by when we add pi/2 to pi (quad 3) and subtract pi/2 (quad 4) from 2pi, we get 3pi/2 both times. I just chose another example to show what to do if there are two unique solutions in that domain.

Notice on the c.a.s.t diagram how the positive x-axis is labelled "0, 2pi".

In quad 1, we add the angle to 0.

In quad 4, we subtract the angle from 2pi. You could just as well subtract it from 0, though, and get a negative angle, which will be equivalent to (2pi - angle) because it will be 2pi out of phase, but if the domain is from 0 to 2pi , we need to add that 2pi to get a solution in the domain.

In quad 2, we subtract the angle from pi.

And in quad 3, we add the angle to pi.

mathfun · Dec 29, 2005

thanx, i'll go try that! thanx thanx thanx everybody!

mathfun · Dec 30, 2005

one last question about this question.

so when i find the quadrants and do the equation, i ignore the negative sign?

so like.. arcsinx= -pi/2....and i want it to be in fourth quadrant(2pi-), i woudl just go 2pi - pi/2 and NOT.. 2pi - (-pi/2)
right? so ignore negative sign, right?

Unco · Dec 30, 2005

Take the arcsin of positive 1.

The negative tells you to be in quadrants 3 and 4.

2pi - pi/2 is correct.

mathfun · Dec 30, 2005

arcsin of positive one is a positive number....

back to my question, ignore the negative sign?

Unco · Dec 30, 2005

The arcsin of a positive number is always positive.

mathfun · Dec 30, 2005

Unco said:
Take the arcsin of positive 1.

The negative tells you to be in quadrants 3 and 4.

2pi - pi/2 is correct.

then what did u mean by that?

take the arcsin of postive 1..... the negative tells me to be in quad III and IV..but there is no negative sin for arsin1

Unco · Dec 30, 2005

sin(x) = -1

We take the arcsine of positive 1 to get pi/2.

The negative in front of 1 tells us to be in quadrants 3 and 4:

As we always do in quadrant 3, we add
x = pi + pi/2 = 3pi/2

As we always do in quadrant 4, we subtract
x = 2pi - pi/2 = 3pi/2

mathfun · Dec 30, 2005

oh ....i just remembered..
shouldn't there be a general answer to this question? as in..
11pi/3 + something something pi? or something?

Unco · Dec 30, 2005

I've lost track of which question you are referring to, but if you are not given a domain and asked for a general solution, then you can apply the formula:

.If \(\displaystyle \L \sin{x} = -1\),

. . . . \(\displaystyle \L x = n\pi \, + \, (-1)^n \cdot (\arcsin(-1))\)

. . . . . . \(\displaystyle \L = n\pi \, + \, (-1)^n(-\frac{\pi}{2})\)

. . . . . . \(\displaystyle \L = n\pi \, - \, (-1)^n(\frac{\pi}{2})\)

. . . . where \(\displaystyle n\) is any integer.

There's no need to worry about the sign; plug'n'chug.

You can use the formula for finding solutions within a domain, of course; it's a matter of preference how you solve trigonometric equations.

mathfun · Jan 2, 2006

would this question have a general answer of not? i'm having trouble to know when there is suppose to be a general answer or not and i'm also having trouble finding the general solution.

Unco said:
mathfun said:

so then it should be:

2(sin(x/2)cos(pi/3) - sin(pi/3)cos(x/2)
=2sin((x/2) - (pi/3))

(x/2) - (pi/3) =

x = (11pi)/6 ??

like that? i understand y we use (3pi)/2...but i don't understnad... y not use -2?
and i never new that for.... sin(alpha - beta) where (alpha - beta) = minimum??

Click to expand...

You have established

. . . \(\displaystyle \L y = 2\sin{(\frac{x}{2} - \frac{\pi}{3})}\)

is minimum at \(\displaystyle y = -2\).

So set \(\displaystyle y = -2\) and solve for \(\displaystyle x\):

. . . \(\displaystyle \L 2\sin{(\frac{x}{2} \, - \, \frac{\pi}{3})} = -2\)

. . . \(\displaystyle \L \sin{(\frac{x}{2} \, - \, \frac{\pi}{3})} = -1\)

Some practise (ie. a long and ugly way):

Sine is negative in quadrants 3 and 4:

Code:

pi/2 s | a | | pi -------+-------- 0, 2pi | /|\ | \/ | \/ / | \ c t | 3pi/2

And \(\displaystyle \L \arcsin{(1)} = \frac{\pi}{2}\)

So, solving for \(\displaystyle \L \frac{x}{2} - \frac{\pi}{3}\) first:

. . . \(\displaystyle \L \begin{align*} \left(\frac{x}{2} \, - \, \frac{\pi}{3}\right) &= \pi \, + \, \frac{\pi}{2} \\
&= \frac{3\pi}{2}\\
\end{align*}\)

. . . (and \(\displaystyle \L \left(\frac{x}{2} - \frac{\pi}{3}\right) = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}\))

Now solve for \(\displaystyle x\):

. . . \(\displaystyle \L \left(\frac{x}{2} \, - \, \frac{\pi}{3}\right) = \frac{3\pi}{2}\)

. . . \(\displaystyle \L \frac{x}{2} = \frac{3\pi}{2} + \frac{\pi}{3}\)

. . . \(\displaystyle \L \frac{x}{2} = \frac{11\pi}{6}\)

. . . \(\displaystyle \L x = \frac{11\pi}{3}\)

If there were two solutions for \(\displaystyle \left(\frac{x}{2} \, - \, \frac{\pi}{3}\right)\), we solve for \(\displaystyle x\) as above for each solution.

Unco · Jan 2, 2006

There's certainly a general solution.

We would hope the question would either give us a domain to work with or tell us to find the general solution.

We have

. . \(\displaystyle \L \sin{\left(\frac{x}{2} \, - \, \frac{\pi}{3}\right)} \, = \, -1\)

Apply the formula:

. . \(\displaystyle \L \frac{x}{2} \, - \, \frac{\pi}{3} \, = \, n\pi \, + \, (-1)^n \cdot \arcsin{(-1)}\)

Solve for \(\displaystyle x\).

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