# trigo identities

• Jul 16th 2007, 04:21 AM
colloquial
trigo identities
solve the eqns for 0<=x<=180

1. sin3x+sinx=0
(i don't get my friend's working):
2 sin2xcosx=0
2(2sinxcosx)(cosx)=0
(4sinxcosx)(cosx)=0
4 sinx=0
sinx=0
x= 0, 180

what I don't get is how the (4 sin x cos x) (cos x) = 0 became just 4 sin x = 0.
she said she canceled it and gave no further explanation. i don't get it, because it's multiply, how could she have canceled the cos x-es out.

2. sin5x-sinx=0
i got up until:
2 cos 3x sin 2x = 0
2 (4cos*3x-3cosx) (2 sinx cosx) = 0
• Jul 16th 2007, 04:43 AM
topsquark
Quote:

Originally Posted by colloquial
solve the eqns for 0<=x<=180

1. sin3x+sinx=0
(i don't get my friend's working):
2 sin2xcosx=0
2(2sinxcosx)(cosx)=0
(4sinxcosx)(cosx)=0
4 sinx=0
sinx=0
x= 0, 180

what I don't get is how the (4 sin x cos x) (cos x) = 0 became just 4 sin x = 0.
she said she canceled it and gave no further explanation. i don't get it, because it's multiply, how could she have canceled the cos x-es out.

I don't know what's going on either.

Is this
$\displaystyle sin^3(x) + sin(x) = 0$
or
$\displaystyle sin(3x) + sin(x) = 0$?

Based on your friend's answer I'm guessing it's the first one. So factor the common sin(x):
$\displaystyle sin(x)(sin^2(x) + 1) = 0$

Thus either $\displaystyle sin(x) = 0$ ==> $\displaystyle x = 0^o, 180^o$
or $\displaystyle sin^2(x) + 1 = 0$ which is valid for no values of x.

So the solution is $\displaystyle 0^o, 180^o$.

-Dan
• Jul 16th 2007, 04:45 AM
colloquial
it's the second one.

eta:
this is what I got -

(4 sin x cos x) (cos x) = 0
cos^2 x . 4 sin x = 0

i.
cos^2x = 0
cos x = 0
x = 90

ii.
4 sin x = 0
sin x = 0
x = 0, 180

therefore, x = 0, 90, 180

is this correct?
• Jul 16th 2007, 06:30 AM
colloquial
anyone can help me with #2? please?
thanks!
• Jul 16th 2007, 07:11 AM
topsquark
Quote:

Originally Posted by colloquial
it's the second one.

eta:
this is what I got -

(4 sin x cos x) (cos x) = 0

Where are you getting this line from?

$\displaystyle sin(3x) + sin(x) = 0$

First:
$\displaystyle sin(3x) = sin(2x + x) = sin(2x)cos(x) + sin(x)cos(2x)$

$\displaystyle = 2sin(x)cos(x) \cdot cos(x) + sin(x)(1 - 2sin^2(x))$

$\displaystyle = 2sin(x)cos^2(x) + sin(x) - 2sin^3(x)$

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

$\displaystyle = 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x)$

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

$\displaystyle [3sin(x) - 4sin^3(x)] + sin(x) = 0$

$\displaystyle -4sin^3(x) + 4sin(x) = 0$

$\displaystyle sin^3(x) - sin(x) = 0$
which looks disturbingly like the other problem... Please check my Math on this!

-Dan
• Jul 16th 2007, 09:48 AM
Soroban
Hello, colloquial!

Quote:

Solve for $\displaystyle 0^o \leq x \leq 180^o$

$\displaystyle 1)\;\;\sin3x + \sin x\:=\:0$

Sum-to-Product identity: .$\displaystyle \sin A + \sin B \:=\:2\!\cdot\!\sin\!\left(\frac{A+B}{2}\right)\!\ cdot\!\cos\!\left(\frac{A-B}{2}\right)$

The equation becomes: .$\displaystyle 2\cdot\sin(2x)\cdot\cos(x)\:=\:0$

We have two equations to solve:

. . $\displaystyle \sin(2x)\,=\,0\quad\Rightarrow\quad 2x \:=\:0^o,\:180^o,\:360^o\quad\Rightarrow\quad x \:=\:0^o,\:90^o,\:180^o$

. . $\displaystyle \cos x \:=\:0\quad\Rightarrow\quad x \,=\,90^o$

Solutions: .$\displaystyle \boxed{x \:=\:0^o,\:90^o,\:180^o}$

Quote:

$\displaystyle 2)\;\;\sin5x - \sin x \:=\:0$
Another identity: .$\displaystyle \sin A - \sin B \:=\:2\!\cdot\!\cos\!\left(\frac{A+B}{2}\right)\!\ cdot\!\sin\!\left(\frac{A-B}{2}\right)$

The equation becomes: .$\displaystyle 2\cdot\cos(3x)\cdot\sin(2x) \:=\:0$

We have two equations to solve:

. . $\displaystyle \cos3x \,=\,0\quad\Rightarrow\quad 3x\:=\:90^o,\:270^o,\:450^o\quad\Rightarrow\quad x \:=\:30^o,\:90^o,\:150^o$

. . $\displaystyle \sin2x \,=\,0\quad\Rightarrow\quad 2x\:=\:0^o,\:180^o,\:360^o\quad\Rightarrow\quad x \:=\:0^o,\:90^o,\:180^o$

Solution: .$\displaystyle \boxed{x\;=\;0^o,\:30^o,\:90^o,\:150^o,\:180^o}$

• Jul 17th 2007, 04:38 AM
topsquark
Quote:

Originally Posted by Soroban
Sum-to-Product identity: .$\displaystyle \sin A + \sin B \:=\:2\!\cdot\!\sin\!\left(\frac{A+B}{2}\right)\!\ cdot\!\cos\!\left(\frac{A-B}{2}\right)$

I always forget that one...

-Dan