sinx+ sin2x+ sin3x= 1
thanks for all

$\displaystyle \displaystyle \sin 2x = 2\sin x \cos x $

$\displaystyle \displaystyle \sin 3x = 3\sin x - 4\sin^3 x $

can you help me complete? it not so easy like that

Originally Posted by quynhnguyen

sinx+ sin2x+ sin3x= 1
thanks for all

You can also prove by induction that:


$\displaystyle \sin(x)+\sin(2x)+\sin(3x)+...+\sin(nx)=\frac{\sin{ (\frac{n}{2}x)}\sin(\frac{n+1}{2}x)}{\sin(\frac{x} {2})}$

please help me complete, : (

Originally Posted by quynhnguyen

please help me complete, : (

Post #2 puts you on the right track. I'll add that line 2 of that reply follows from using the compound angle formula on sin(x + 2x).

You need to show some effort if more help is needed. What have you tried and where are you still stuck? Are you expected to solve it by hand or using CAS technology?

Hint!

I will post the whole solution soon, but take a moment and think why multiplying the both sides of an equation with $\displaystyle 2\sin{\frac{x}{2}}$ will help you...

Not the whole solution...

With help of my formula we get:

$\displaystyle \sin2x\sin{\frac{3x}{2}}=\sin{\frac{x}{2}}$

$\displaystyle 2\sin{x}\cos{x}\sin{\frac{3x}{2}}=\sin{\frac{x}{2} }$

$\displaystyle 4\sin {\frac{x}{2}} \cos{\frac{x}{2}} \cos{x} \sin {\frac{3x}{2}}=\sin{\frac{x}{2}}$



$\displaystyle \sin {\frac{x}{2}} (4\cos {\frac{x}{2}} \cos {x} \sin {\frac{3x}{2}}-1)=0$


$\displaystyle \sin {\frac{x}{2}} (4\cos {\frac{x}{2}}\cos {x} \sin {\frac{3x}{2}}-1)=0$

$\displaystyle \sin {\frac{x}{2}} (2\sin {2x}+\sin {2x}-1)=0$

$\displaystyle \sin {\frac{x}{2}} (2 \sin {2x} \cos {x}+\sin {2x}-1)=0$

Originally Posted by Also sprach Zarathustra

Hint!

I will post the whole solution soon, but take a moment and think why multiplying the both sides of an equation with $\displaystyle 2\sin{\frac{x}{2}}$ will help you...

Are you sure there's no typo? Maybe the equation is:
$\displaystyle \sin(x)+\sin(2x)+\sin(3x)=0$
?

you can use the identity
$\displaystyle \sin{P}+\sin{Q}=2\sin(\frac{P+Q}{2})\cos(\frac{P-Q}{2})$
using arrangement $\displaystyle \sin{3x}+\sin{x}+\sin{2x}$

Originally Posted by waqarhaider

you can use the identity
$\displaystyle \sin{P}+\sin{Q}=2\sin(\frac{P+Q}{2})\cos(\frac{P-Q}{2})$
using arrangement $\displaystyle \sin{3x}+\sin{x}+\sin{2x}$

thanks but it not so easy like that

Originally Posted by Siron

Are you sure there's no typo? Maybe the equation is:
$\displaystyle \sin(x)+\sin(2x)+\sin(3x)=0$
?

yes i'm sure: sinx+ sin2x+ sin3x= 1
if: sinx+ sin2x+ sin3x= 0 it too easy thats right? so i needn't post this trigonometri in here

Originally Posted by quynhnguyen

sinx+ sin2x+ sin3x= 1
thanks for all

This equation does not seem to have a clean solution. If you graph the function, you can see that there are two solutions between 0º and 90º (and no others in the range 0º to 360º). The smaller solution is approximately 9.836º, but that does not look like a recognisable quantity in either degrees or radians. I haven't tried to locate the other solution accurately, but it is somewhere near 71.3º. Again, this is not a nice-looking angle. The solutions for $\displaystyle \sin x$ are not rational. I doubt whether the equation can be solved exactly.

Originally Posted by Opalg

This equation does not seem to have a clean solution. If you graph the function, you can see that there are two solutions between 0º and 90º (and no others in the range 0º to 360º). The smaller solution is approximately 9.836º, but that does not look like a recognisable quantity in either degrees or radians. I haven't tried to locate the other solution accurately, but it is somewhere near 71.3º. Again, this is not a nice-looking angle. The solutions for $\displaystyle \sin x$ are not rational. I doubt whether the equation can be solved exactly.

yes . Here: $\displaystyle \ sinx + sin2x + sin3x = 1$ graph