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Trigonometry Trigonometry questions I'm stuck on

nicodemus

New member
Jul 22, 2012
16
Good Day,

I'm stuck on two trigonometry questions.

One requires me to prove a trigonometry summation and the other requires me to find the general solution of the given equation.

1. I am required to prove the b/m
sin x + sin 2x + sin 3x + ... + sin nx = {sin (n/2)x sin [(n+1)/2]x}/ sin (x/2)

I've attempted to prove this using mathematical induction. However, I'm stuck and need help/ advice on how I should proceed further. Attached is the work I have done so far.

View attachment 279

2. I have to find the general solution of the following equation:
tan (2x - pi) = sin x

I've been able to get till tan 2x = sin x. However, whatever I've done after that doesn't give me the correct solution, which is
x = n*pi or 2n*pi + 1.946
.

Your help/ advice will be greatly appreciated.

Thanks in advance.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Good Day,

I'm stuck on two trigonometry questions.

One requires me to prove a trigonometry summation and the other requires me to find the general solution of the given equation.

1. I am required to prove the b/m
sin x + sin 2x + sin 3x + ... + sin nx = {sin (n/2)x sin [(n+1)/2]x}/ sin (x/2)

I've attempted to prove this using mathematical induction. However, I'm stuck and need help/ advice on how I should proceed further. Attached is the work I have done so far.

View attachment 279

2. I have to find the general solution of the following equation:
tan (2x - pi) = sin x

I've been able to get till tan 2x = sin x. However, whatever I've done after that doesn't give me the correct solution, which is
x = n*pi or 2n*pi + 1.946
.

Your help/ advice will be greatly appreciated.

Thanks in advance.
Hi nicodemus, :)

1) Use the double angle formula for \(\sin(k+1)\theta\). Then you will get,

\[\sin x + \sin 2x + \sin 3x + ... + \sin kx+\sin(k+1)x=\sin\left(\frac{k+1}{2}\right)\theta\left[\frac{\sin\frac{k\theta}{2}+2\sin\frac{\theta}{2}\cos\left(\frac{k+1}{2}\right)\theta}{\sin\frac{ \theta}{2}}\right]\]

Now use the angle sum identity for \(\cos\left(\frac{k+1}{2}\right)\theta\) and simplify(Using Double angle formulas of Sine and Cosine) to obtain,

\[\sin x + \sin 2x + \sin 3x + ... + \sin kx+\sin(k+1)x=\sin\left(\frac{k+1}{2}\right)\theta\left[\frac{\sin\frac{k\theta}{2}\cos\theta+\sin\theta \cos\frac{k\theta}{2}}{\sin\frac{ \theta}{2}}\right]\]

Using the angle sum identity again, \(\sin\left(\frac{k+2}{2}\right)\theta=\sin\frac{k\theta}{2}\cos\theta+\sin\theta\cos\frac{k\theta}{2}\) you can get the required answer.

2) \[\tan2x=\sin x\]

\[\Rightarrow \frac{2\sin x\cos x}{\cos 2x}=\sin x\]

\[\Rightarrow \sin x=0\mbox{ or }2\cos x-\cos 2x=0\]

\[\Rightarrow \sin x=0\mbox{ or }2\cos^{2}x-2\cos x-1=0\]

Hope you can continue. :)

Kind Regards,
Sudharaka.
 

nicodemus

New member
Jul 22, 2012
16
Good Day,

Thank you so much for your help.

However, I don't understand how you got this part:

[FONT=MathJax_Main]⇒[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main] or [/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]​

Could you please give a more detailed explanation?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Good Day,

Thank you so much for your help.

However, I don't understand how you got this part:

[FONT=MathJax_Main]⇒[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main] or [/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]

Could you please give a more detailed explanation?
You are welcome. :)

\[\frac{2\sin x\cos x}{\cos 2x}=\sin x\]

\[\Rightarrow \sin x\left(\frac{2\cos x}{\cos 2x}-1\right)=0\]

\[\Rightarrow \sin x\left(\frac{2\cos x-\cos 2x}{\cos 2x}\right)=0\]

Since, \(\frac{1}{\cos 2x}\neq 0\) we have,

\[\sin x(2\cos x-\cos 2x)=0\]

\[\therefore \sin x=0\mbox{ or }2\cos x-\cos 2x=0\]

Is it clear to you now? :)
 

nicodemus

New member
Jul 22, 2012
16
Brilliant! It's crystal clear to me now.

Thanks very much!
 

nicodemus

New member
Jul 22, 2012
16
Good Day,

I'm sorry about this but I'm still having trouble finding the general solution for this part


[FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]


I applied the Quadratic Formula and got cos x = (1 + 3) / 2 but I don't think it's anywhere near this: 2n*pi + 1.946.

What am I doing wrong?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Good Day,

I'm sorry about this but I'm still having trouble finding the general solution for this part


[FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]


I applied the Quadratic Formula and got cos x = (1 + 3) / 2 but I don't think it's anywhere near this: 2n*pi + 1.946.

What am I doing wrong?
Hi nicodemus, :)

The answer you have obtained for \(\cos x\) is correct. However you should solve for \(x\). First see whether both of these answers are possible for a cosine value. Note that \(cos x\leq 1\).

Kind Regards,
Sudharaka.