Trigonometry Trigonometry questions I'm stuck on

[nicodemus]

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]

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]

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]

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]

Brilliant! It's crystal clear to me now.

Thanks very much!

 

[nicodemus]

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]

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.