Minimum Value

[mathfun]

The minimum value of sin(x/2) - squareroot(3)cos(x/2) is attained when x is ?

I totally don't even know where to start.

 

[tkhunny]

Have you met the First Derivative? You may not need it, but it would help.

Electronics folks know a little trick for amplitude:

a*sin(variable) + b*cos(variable) ==> Total Amplitude = sqrt(a^2 + b^2)

It should look a lot like the Pythagorean Theorem.

In your case, sqrt(1^2 + sqrt(3)^2) = sqrt(1+3) = sqrt(4) = 2

This makes:
Maximum Value = 2
Minimum Value = -2

 

[mathfun]

Have you met the First Derivative? You may not need it, but it would help.

what are derivatives? what are first derivative? I have not learn about them..yet.

Electronics folks know a little trick for amplitude:

a*sin(variable) + b*cos(variable) ==> Total Amplitude = sqrt(a^2 + b^2)

amplitude? can you explain this a bit further, please?

 

[tkhunny]

OK, forget about the derivative. I forgot you weren't up to that, yet. No worries. you'll get there soon enough.

"Amplitude" is a characteristic of your basic trig functions. sin(x) hangs around [-1,1]. It's amplitude is 1. 2*sin(x) hangs around [-2,2]. It's amplitude is 2. That's all it is, just a way to describe some of the features.

A good use is for such a problem as this: Solve sin(x) = 12. Answer: sin(x) doesn't get any bigger than 1, so it can't be 12.

In this case, you may want to pull a sneaky trick.

If f(x) = sin(x/2) - sqrt(3)*cos(x/2),
then f(x)/2 = (1/2)*sin(x/2) - (sqrt(3)/2)*cos(x/2)
= sin(x/2)*cos(pi/3) - cos(x/2)*sin(pi/3)
= sin(x/2 - pi/3),
making f(x) = 2*sin(x/2 - pi/3) and it's a little easier to see.

 

[mathfun]

If f(x) = sin(x/2) - sqrt(3)*cos(x/2)

so u mean..we put everything over 2... and then..
then f(x)/2 = (1/2)*sin(x/2) - (sqrt(3)/2)*cos(x/2)
= sin(x/2)*cos(pi/3) - cos(x/2)*sin(pi/3)

i can't follow how it turns into that. where did pi/3 come from?

making f(x) = 2*sin(x/2 - pi/3) and it's a little easier to see.

what do u mean making it a little easier to see for what?
i don't understand what we are doing. we are trying to find the minimum of that equation... how do you do that?

 

[Gene]

1/2 = cos(pi/3)
What he did was make it fit the difference of two angles formula. sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
Then he used it backwards to find sin(a-b).
As they said, sin(anything) has a minimum of -1 so 2sin(anything) has a minimum of -2

 

[tkhunny]

I may just be barking up excessively dark trees.

I think my most important point is that there are many ways to solve many things. Feel free to pick one you understand, but don't get caught in the trap of thinking there is only one way. You will need all the tools you can get your hands on.

 

[mathfun]

ohh ok.. but i still can't see how this works


f(x)/2 = (1/2)*sin(x/2) - (sqrt(3)/2)*cos(x/2)
= sin(x/2)*cos(pi/3) - cos(x/2)*sin(pi/3)

where did the constants go?

 

[Gene]

I told you a couple posts ago that
1/2 = cos(pi/3)
I didn't mention it but...
sqrt(3)/2 = sin(pi/3)
He wanted a trig function so he substituted cos(pi/3) for 1/2 and
sin(pi/3) for sqrt(3)/2
That's where the constants went to.
----------------
Gene

 

[mathfun]

OHHHH... ic. thanx
omgosh..so complicated! i would have NEVER thot of using this way... how did u think of this?!?!
so the answer is -2?

 

[tkhunny]

What are you still questioning? The answer has come out -2 three times already. Is there a methodology that you can use to obtain this result? Maybe nothing shown here was of any use to you.

 

[Gene]

Yes, it is. That is a rather tricky way of using the sum of two angles formula to solve the problem. He knows the formulas and saw that it was close to that one and saw a way to change it so it was exactly the same.
-----------------
Gene

 

[mathfun]

What are you still questioning?

i am still questioning because the answer is not correct, i'm afraid to say. i think it is because it is not asking of the minimum of the question, but what x is.

the question once again, is: The minimum value of sin(x/2) - squareroot(3)cos(x/2) is attained when x is ?

 

[mathfun]

Is there a methodology that you can use to obtain this result? Maybe nothing shown here was of any use to you.

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.

 

[stapel]

Using the identity suggested earlier:

. . .sin(x/2) - sqrt[3]cos(x/2)

. . . . .= 1×sin(x/2) - sqrt[3] cos(x/2)

. . . . .= (2/1)(1/2) [1×sin(x/2) - sqrt[3] cos(x/2)]

. . . . .= 2 [(1/2) sin(x/2) - [sqrt[3]/2] cos(x/2)]

. . . . .= 2 [cos(30°) sin(x/2) - sin(30°) cos(x/2)]

. . . . .= 2 sin(x/2 - 30°)

This will take on its minimum value (of negative two, as noted) when the sine curve is at its minimum. The sine takes on its minimum when the argument is \(\displaystyle \theta\,=\,\frac{3\pi}{2}\), or at 270° (or appropriate repeats, of course). So:

. . . . .x/2 - 30° = 270°

. . . . .x/2 = 300°

. . . . .x = 600°

Check my work! :wink:

Eliz.

 

[Unco]

2 [(1/2) sin(x/2) - [sqrt[3]/2] cos(x/2)]

= 2 [cos(30°) sin(x/2) - sin(30°) cos(x/2)]

\(\displaystyle \L \cos{30} = \frac{\sqrt{3}}{2}\) and \(\displaystyle \L \sin{30} = \frac{1}{2}\)

 

[mathfun]

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??

 

[stapel]

Nuts! Did I reverse those...again? Okay....

. . .sin(x/2) - sqrt[3]cos(x/2)

. . . . .= 1×sin(x/2) - sqrt[3] cos(x/2)

. . . . .= (2/1)(1/2) [1×sin(x/2) - sqrt[3] cos(x/2)]

. . . . .= 2 [(1/2) sin(x/2) - [sqrt[3]/2] cos(x/2)]

. . . . .= 2 [sin(30°) sin(x/2) - cos(30°) cos(x/2)]

. . . . .= 2 (-cos(x/2 - 30°))

Well, the minimum value is still -2, but it will occur at the maximum for the cosine curve, which would be when the argument equals zero, so:

. . . . .x/2 - 30° = 0°

. . . . .x/2 = 30°

. . . . .x = 60°

...maybe....

Eliz.

 

[Unco]

2 [sin(30°) sin(x/2) - cos(30°) cos(x/2)]

= 2 (-cos(x/2 - 30°))

sin(30°) sin(x/2) - cos(30°) cos(x/2)
= - [cos(30°) cos(x/2) - sin(30°) sin(x/2)]
= - [cos(x/2 + 30°)]

 

[stapel]

D'oh!

Definitely time for me to back awaaaaaaay from the keyboard....

Eliz.