Hard question on Mathematical Induction and Trigonometry Identity

[Masaru]

I have difficulty working out the following question:

Use the principle of mathematical induction to prove that:

cos^2(x) + cos^2(2x) + cos^2(3x) + cos^2(4x) +.....+cos^2(nx) = (1/2)[n +(cos[(n + 1)x]sin(nx)]/sinx] for all positive integers n.

My working is as follows:

Proposition P_n is above.

1) If n = 1, LHS = cos^2(x), and RHS = (1/2)(1 + (cos2xsinx/sinx) = (1/2)(1 + cos(2x)) = (1/2)(1 + 2cos^2(x) - 1) = cos^2(x). Therefore, P₁ is true.

2) If P_k is true, then

cos^2(x) + cos^2(2x) + cos^2(3x) + cos^2(4x) +.....+cos^2(kx) = (1/2)[k +(cos[(k + 1)x]sin(kx)]/sinx] for all positive integers k.​

Now P_K+1 is:

{cos^2(x) + cos^2(2x) + cos^2(3x) + cos^2(4x) +.....+cos^2(kx)}+

cos^2[(k+1)x] =

​

(1/2)[k +(cos[(k + 1)x]sin(kx))/sinx] +

cos^2[(k+1)x] =

(1/2)[k +(cos[(k + 1)x]sin(kx))/sinx] + (1/2)(cos[2(k+1)x] + 1) <Using trig identity cos2A = 2cos^2(A) - 1)

= (1/2)[(k + 1) +

(cos[(k + 1)x]sin(kx))/sinx] + cos

[2(k+1)x]]​

​

​

= (1/2)

[(k + 1) +

(cos[(k + 1)x]sin(kx) + sin(x)

cos[2(k+1)x]​

)/sinx]

And I am stuck with no idea how I can change the last line above to

(1/2)[(k + 1) +

(cos[((k + 1)+1)x]sin[(k+1)x/sinx] so I can say that P_k+1 is true whenever P_k is true because it is so complex and difficult for me.

I would much appreciate it if someone can help me with this.

P.S. My apology for using so many brackets as above - I do not know how to make it display on the screen just like the way we write with hands. So if someone knows how to do it and let me know, that would be much appreciated as well.

Thank you.​

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[Steven G]

I have difficulty working out the following question:

Use the principle of mathematical induction to prove that:

cos^2(x) + cos^2(2x) + cos^2(3x) + cos^2(4x) +.....+cos^2(nx) = (1/2)[n +(cos[(n + 1)x]sin(nx)]/sinx] for all positive integers n.

My working is as follows:

Proposition P_n is above.

1) If n = 1, LHS = cos^2(x), and RHS = (1/2)(1 + (cos2xsinx/sinx) = (1/2)(1 + cos(2x)) = (1/2)(1 + 2cos^2(x) - 1) = cos^2(x). Therefore, P₁ is true.

2) If P_k is true, then

cos^2(x) + cos^2(2x) + cos^2(3x) + cos^2(4x) +.....+cos^2(kx) = (1/2)[k +(cos[(k + 1)x]sin(kx)]/sinx] for all positive integers k.​

Now P_K+1 is:

{cos^2(x) + cos^2(2x) + cos^2(3x) + cos^2(4x) +.....+cos^2(kx)}+

cos^2[(k+1)x] =

​

(1/2)[k +(cos[(k + 1)x]sin(kx))/sinx] +

cos^2[(k+1)x] =

(1/2)[k +(cos[(k + 1)x]sin(kx))/sinx] + (1/2)(cos[2(k+1)x] + 1) <Using trig identity cos2A = 2cos^2(A) - 1)

= (1/2)[(k + 1) +

(cos[(k + 1)x]sin(kx))/sinx] + cos

[2(k+1)x]]​

​

​

= (1/2)

[(k + 1) +

(cos[(k + 1)x]sin(kx) + sin(x)

cos[2(k+1)x]​

)/sinx]

And I am stuck with no idea how I can change the last line above to

(1/2)[(k + 1) +

(cos[((k + 1)+1)x]sin[(k+1)x/sinx] so I can say that P_k+1 is true whenever P_k is true because it is so complex and difficult for me.

I would much appreciate it if someone can help me with this.

P.S. My apology for using so many brackets as above - I do not know how to make it display on the screen just like the way we write with hands. So if someone knows how to do it and let me know, that would be much appreciated as well.

Thank you.​

​

​

​

​

​

This equation cos^2(x) + cos^2(2x) + cos^2(3x) + cos^2(4x) +.....+cos^2(nx) = (1/2)[n +(cos[(n + 1)x]sin(nx)]/sinx] does not match If n = 1, LHS = cos^2(x), and RHS = (1/2)(1 + (cos2xsinx/sinx) = (1/2)(1 + cos(2x)) = (1/2)(1 + 2cos^2(x) - 1) = cos^2(x).

You opened two brackets like [ and closed three of them!! Fix this and I'll look at again.

 

[pka]

I have difficulty working out the following question:
Use the principle of mathematical induction to prove that:
cos^2(x) + cos^2(2x) + cos^2(3x) + cos^2(4x) +.....+cos^2(nx) = (1/2)[n +(cos[(n + 1)x]sin(nx)]/sinx] for all positive integers n.

It may help to rewrite as: $\displaystyle \sum\limits_{j = 1}^K {{{\cos }^2}(jx)} = RHS(K)$That notation allows simplification, so $\displaystyle \sum\limits_{j = 1}^{K + 1} {{{\cos }^2}(jx)} = \sum\limits_{j = 1}^K {{{\cos }^2}(jx)} + {\cos ^2}([K + 1]x)$


BUT I absolutely agree with Jomo the RHS is a total unreadable mess!

 

[Masaru]

Resubmission of the question.

This equation cos^2(x) + cos^2(2x) + cos^2(3x) + cos^2(4x) +.....+cos^2(nx) = (1/2)[n +(cos[(n + 1)x]sin(nx)]/sinx] does not match If n = 1, LHS = cos^2(x), and RHS = (1/2)(1 + (cos2xsinx/sinx) = (1/2)(1 + cos(2x)) = (1/2)(1 + 2cos^2(x) - 1) = cos^2(x).

You opened two brackets like [ and closed three of them!! Fix this and I'll look at again.

What I put did not make sense because inside each bracket on LHS is not multiplied by 2 but what I wanted mean is like (cos (...))^2. This equation is very difficult to read if you try to express it with ^ , brackets and slash, and that is the reason why I asked you how I can display mathamatical symbols in HTML because I have no idea how to do it. I went to some websites like https://www.mathjax.org/ and https://www.mathcha.io, but I cannot figure it out I am still unable to display equations in the same way as handwriting.

Anyway, since I do not know how to display and express the equation properly in HTML here, I have typed out my question and all of my workings in Word document on PC screeen, took photos and attached them here so you can perfectly understand the question and my workings.

So please have a look at the attachment here.

 

[Steven G]

What I put did not make sense because inside each bracket on LHS is not multiplied by 2 but what I wanted mean is like (cos (...))^2. This equation is very difficult to read if you try to express it with ^ , brackets and slash, and that is the reason why I asked you how I can display mathamatical symbols in HTML because I have no idea how to do it. I went to some websites like https://www.mathjax.org/ and https://www.mathcha.io, but I cannot figure it out I am still unable to display equations in the same way as handwriting.

Anyway, since I do not know how to display and express the equation properly in HTML here, I have typed out my question and all of my workings in Word document on PC screeen, took photos and attached them here so you can perfectly understand the question and my workings.

So please have a look at the attachment here.

Some things cancel out. That is you need to show that:
cos[(k + 1)x]sin(kx) + sin(x)cos(2(k+1)x)= cos((k + 2)x)sin(k+1)x

What have you tried? Use some identities to see if it helps. I would first pick a value for k and x to see if both sides are equal as it is easy to make a mistake with this mess.

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[Masaru]

Still hard and I am stuck again.

Some things cancel out. That is you need to show that:
cos[(k + 1)x]sin(kx) + sin(x)cos(2(k+1)x)= cos((k + 2)x)sin(k+1)x

What have you tried? Use some identities to see if it helps. I would first pick a value for k and x to see if both sides are equal as it is easy to make a mistake with this mess.

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Please open the attachment to see my working.
I am stuck again and I do not think that I am getting any closer to the RHS.

 

[Harry_the_cat]

I'd start with the RHS of your proof for now and use expansions for cos(A+B) and sin (A+B).

 

[Masaru]

Thank you.

I'd start with the RHS of your proof for now and use expansions for cos(A+B) and sin (A+B).

Thank you for your advice.

Finally I worked it out but I have ended up with more than 2 pages of long workings.

 

[Denis]

Finally I worked it out but I have ended up with more than 2 pages of long workings.

Well, on the bright side: that's less than 3

 

[Steven G]

Well, on the bright side: that's less than 3

You are getting there! One day you'll be able to count to 10. You have come so far! Everyone on the forum is so proud of you.

 

[Steven G]

Thank you for your advice.

Finally I worked it out but I have ended up with more than 2 pages of long workings.

Welcome to the beginning of real math. There are some mathematicians who have spent their whole career working on a single problem and never solving it!

 

[Denis]

You are getting there! One day you'll be able to count to 10. You have come so far! Everyone on the forum is so proud of you.

Ahhhh Jomo...everything I know I learned from you...time for a change!!