1. Trignometric Question

Hi,
Can you help me with the following question please?

3sinx = sin3x

My working:
3sinx - sin3x = 0
3sinx - sin(x + 2x) = 0
3 sinx - sinx - sin2x = 0
2sinx - 2sinxcosx =0
2sinx ( 1 - cos x) = 0

2 sinx = 0 x= 0 x = 0 and 360 degrees
1 - cos x = 0 cos x = 1 x = 0 and 360 degrees.

Can you help I think I have missed a value between 0 and 360?

Thanks
Cromlix

2. Originally Posted by cromlix
Hi,
Can you help me with the following question please?

3sinx = sin3x

My working:
3sinx - sin3x = 0
3sinx - sin(x + 2x) = 0
3 sinx - sinx - sin2x = 0
2sinx - 2sinxcosx =0
2sinx ( 1 - cos x) = 0

2 sinx = 0 x= 0 x = 0 and 360 degrees
1 - cos x = 0 cos x = 1 x = 0 and 360 degrees.

Can you help I think I have missed a value between 0 and 360?

Thanks
Cromlix
$\displaystyle sin(x+2x) \neq sin(x)+sin(2x)$

try to use $\displaystyle sin(a+b)=sin (a) cos (b)+cos (a) sin (b)$
to expand $\displaystyle sin (x+2x)$

--hope it'll help--

3. $\displaystyle 3\sin{\left(x\right)} = \sin{\left(3x\right)}$

$\displaystyle 3\sin{\left(x\right)} = \sin{\left(x\right)}\left[2\cos{\left(2x\right)}+1\right]$

$\displaystyle 3 = 2 \cos{\left(2x\right)} + 1$

$\displaystyle 1 = \cos{\left(2x\right)}$

$\displaystyle \cos^{-1}{\left(1\right) = 2x$

$\displaystyle 0 = 2x$

$\displaystyle x = 0$ or $\displaystyle x = \pi n$

takes me forever to do latex...

4. Hello, cromlix!

We know the basic identities:

. . $\displaystyle \sin(A + B) \:=\:\sin A\cos B + \cos A\sin B$

. . $\displaystyle \sin2A \:=\:2\sin A\cos A$

. . $\displaystyle \cos2A \:=\:1 - 2\sin^2\!A \:=\:2\cos^2\!A - 1 \:=\:\cos^2\!A - \sin^2\!A$

$\displaystyle \text{Solve for }x\!:\;\;3\sin x \:=\: \sin3x$ .[1]

We need a Triple-Angle Identity . . .

$\displaystyle \sin3x \;=\;\sin(x+2x)$

. . . . . $\displaystyle =\;\sin x\cos2x + \cos x\sin2x$

. . . . . $\displaystyle =\;\sin x(1 - 2\sin^2\!x) + \cos x(2\sin x\cos x)$

. . . . . $\displaystyle =\; \sin x - 2\sin^3\!x + 2\sin x\cos^2\!x$

. . . . . $\displaystyle =\; \sin x - 2\sin^3\!x + 2\sin x(1 - \sin^2\!x)$

. . . . . $\displaystyle =\;\sin x - 2\sin^3\!x + 2\sin x - 2\sin^3\!x$

$\displaystyle \sin3x \;=\; 3\sin x - 4\sin^3\!x$

Equation [1] become: .$\displaystyle 3\sin x \;=\;3\sin x - 4\sin^3\!x$

. . $\displaystyle 4\sin^3\!x \:=\:0 \quad\Rightarrow\quad \sin x \:=\:0$

Therefore: .$\displaystyle x \:=\:\pi n\;\text{ for any integer }n$

5. Trigonometric question

Thank you all very much for your help.

Cromlix