Mean value theorem for trigonometric function

[Raghav Gupta]

Homework Statement

Verify Lagrange's MVT for
$f(x)= sinx - sin2x$ in [ 0, π ]

Homework Equations

$f'(c) = \frac{f(b)-f(a)}{b-a}$

The Attempt at a Solution

Got on solving cosx= 2cos2x
How to find c lies in [0, π ]?
Solved it using quadratic equation but it gives a complicated value inside arccos.

 

[AdityaDev]

On using that formula you get dy/dx at c = 0
dy/dx = cosx-2cos2x now to verify MVT,dy/dx has to be equal to the value obtained from your equation for atleast one c within 0 to pi.
putting cosx=2cos2x, you are indirectly finding the values of that c. If cosx can never be equal to 2cos2x in the given interval, then mvt is invalid here. Else MVT is valid.

 

[SammyS]

Homework Statement

Verify Lagrange's MVT for
$f(x)= sinx - sin2x$ in [ 0, π ]

Homework Equations

$f'(c) = \frac{f(b)-f(a)}{b-a}$

The Attempt at a Solution

Got on solving cosx= 2cos2x
How to find c lies in [0, π ]?
Solved it using quadratic equation but it gives a complicated value inside arccos.

Yes, the value inside the arccosine contains √(33) .

 

[Raghav Gupta]

Yes, the value inside the arccosine contains √(33) .

No it contains (1± √(33))/8.
How to determine that it's in interval [0, π] ?

 

[SammyS]

No it contains (1± √(33))/8.
How to determine that it's in interval [0, π] ?

Well that does contain √(33) .

arccos( y₀) always gives a value in the interval [0, π] , provided that y₀ is in the interval [-1, 1], the domain of the arccos function.

 

[Raghav Gupta]

Well that does contain √(33) .

But
Cosx=2cos2x
=> cosx= 2( 2cos²x - 1)
=> cosx = 4cos²x - 2
=> cosx = (-1 ± √(33))/8
So there is a contradiction with your statement?

 

[SammyS]

But
Cosx=2cos2x
=> cosx= 2( 2cos²x - 1)
=> cosx = 4cos²x - 2
=> cosx = (-1 ± √(33))/8
So there is a contradiction with your statement?

I see √(33) right there in (-1 ± √(33))/8 .

I just said it contains it.

 

[Raghav Gupta]

I see √(33) right there in (-1 ± √(33))/8 .

I just said it contains it.

Sorry__ Sorry. I thought you meant that x= arccos(√(33)).
My mistake.
Thanks the major contribution from your side was telling me domain for arccos and range.

 

[Ray Vickson]

Sorry__ Sorry. I thought you meant that x= arccos(√(33)).
My mistake.
But here we are getting 2 values which are in domain of arccos and hence in the interval.
But shouldn't that be one value only.

Have you tried plotting the two functions over $[0, \pi]$?

 

[Raghav Gupta]

Have you tried plotting the two functions over $[0, \pi]$?

See the plot
What does that mean?

 

[Ray Vickson]

See the plot
What does that mean?

Draw the graphs of $y =2 \cos(2x)$ and $y = \cos(x)$ for $x \in [0,\pi]$, or perhaps the single graph of $y = 2 \cos(2x) - \cos(x)$.

 

[Raghav Gupta]

Draw the graphs of $y =2 \cos(2x)$ and $y = \cos(x)$ for $x \in [0,\pi]$, or perhaps the single graph of $y = 2 \cos(2x) - \cos(x)$.

Here it is newplot.
Then what?

 

[Ray Vickson]

Here it is newplot.
Then what?

Then look at it and think about what it is telling you.

 

[AdityaDev]

Here it is newplot.
Then what?

you use wolftam's graph calculator? Draw it yourself on a paper. Dont depend on graph calculators.
see where the curves intesect. Those are the points where f'(c) is zero.

 

[Raghav Gupta]

But wolfram graph calculator gives correct results. I was using that here because it gives correct sketch and by hand we would get a rough sketch as drawing graphs of complicated functions could be difficult like
sinx-sin2x.

Then look at it and think about what it is telling you.

I think drawing only the graph of sinx-sin2x was okay?
By seeing it the average slope is 0.
Now according to MVT there is or are points on the function in given interval where the slopes of it equal average slope. Seeing the graph we see two points ( one is local minima and other is local maxima).

 

[Raghav Gupta]

By the way thanks Aditya, I saw 2 points where curves intersect and that means we get two values of c for f'(c) =0 .
Also thank you Mr. Ray as I see two points where the curve cosx- 2cos2x cuts the x-axis meaning we get two values of c for f'(c)=0.