# TrigonometryTrigonometry questions I'm stuck on

#### nicodemus

##### New member
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
.

#### Sudharaka

##### Well-known member
MHB Math Helper
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
.

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
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
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
Brilliant! It's crystal clear to me now.

Thanks very much!

#### nicodemus

##### New member
Good Day,

[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
Good Day,

[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.