a tough one

[allegansveritatem]

Here is the problem:

I am dealing in this post with part a). I was petty much at sea here and came up with several feints at a solution. Here is the best of the (false) conclusions I came to:

I'm not sure what is being asked for. Does the problem require that sin^2 omega t be transformed into something like c+a cos(bt)?

 

[Steven G]

[math]\sin(2x)\neq 2sin(2x)(cos(2x))[/math]

 

[Dr.Peterson]

Look at the double-angle formulas for cosine, not sine. You'll see that one of them involves sin^2; solve it for sin^2 and substitute. (I'm intentionally leaving out what the argument will be.)

 

[allegansveritatem]

[math]\sin(2x)\neq 2sin(2x)(cos(2x))[/math]

right...sin2x=2sinxcosx. I see I went astray.

 

[HallsofIvy]

The "average value" of such a function, y= f(x), from x= a to x= b, is given by \(\displaystyle \frac{\int_a^b f(x) dx}{b- a}\).

 

[Deleted member 4993]

Here is the problem:View attachment 18674
View attachment 18675
I am dealing in this post with part a). I was petty much at sea here and came up with several feints at a solution. Here is the best of the (false) conclusions I came to:
View attachment 18676
I'm not sure what is being asked for. Does the problem require that sin^2 omega t be transformed into something like c+a cos(bt)?

For which question/s are you writing (1) & (2) & (3)?

What are the differences - if any - between (1) & (2) & (3)?

How did (2 * pi/w) get in those expressions?

 

[allegansveritatem]

Look at the double-angle formulas for cosine, not sine. You'll see that one of them involves sin^2; solve it for sin^2 and substitute. (I'm intentionally leaving out what the argument will be.)

cos2x=1-2sin^2x...solve what for sin^2? Solve c +a cosbt? Well, I will try that and post back. Thanks.

Look at the double-angle formulas for cosine, not sine. You'll see that one of them involves sin^2; solve it for sin^2 and substitute. (I'm intentionally leaving out what the argument will be.)

I worked on this today and on the third try I got this:


Compared to the results of my earlier endeavors this looks at least hopeful.

 

[allegansveritatem]

For which question/s are you writing (1) & (2) & (3)?

What are the differences - if any - between (1) & (2) & (3)?

How did (2 * pi/w) get in those expressions?

2 pi/b is my attempt to represent the period.

 

[allegansveritatem]

The "average value" of such a function, y= f(x), from x= a to x= b, is given by \(\displaystyle \frac{\int_a^b f(x) dx}{b- a}\).

I thought this: Since c=0, (see the diagram) therefore the average is 0.