trigo: proving an identity

**cakeboby** · December 28th 2009, 09:58 AM

Prove the following identities:
$\frac {sin3a}{sina} - \frac {sin3b}{sinb}= 4sin(b+a)sin(b-a)$

Here's what I did:
L.H.S.
$=3-4sin^2a-(3-4sin^2b)$
$=4(sin^2 b - sin^2 a)$
???

Thanks in advance.

**Soroban** · December 28th 2009, 10:55 AM

Hello, cakeboby!

Prove the following identity: . $\frac {\sin3a}{\sin a} - \frac {\sin3b}{\sin b}\:=\: 4\,\sin(b+a)\,\sin(b-a)$

Here's what I did:

$\text{L.H.S.} \:=\:(3-4\sin^2a)-(3-4\sin^2b)$

. . . . . $=\;4(\sin^2b - \sin^2a) \qquad\hdots \;\;??$

You're doing great!

We need some Sum-to-Product idenities:

. . $\begin{array}{ccc}\sin X + \sin Y &=& 2\sin\left(\dfrac{X+Y}{2}\right)\cos\left(\dfrac{X-Y}{2}\right) \\ \\[-3mm]<br /> \sin X - \sin Y &=& 2\cos\left(\dfrac{X+Y}{2}\right)\sin\left(\dfrac{X-Y}{2}\right) \end{array}$

You have: . $4(\sin a + \sin b)(\sin a - \sin b)$

. . . . . . . $= \;4\bigg[2\sin\left(\frac{b+a}{2}\right)\cos\left(\frac{b-a}{2}\right)\bigg]\,\bigg[2\cos\left(\frac{b+a}{2}\right)\sin\left(\frac{b-a}{2}\right)\bigg]$

. . . . . . . $= \;4\underbrace{\bigg[2\sin\left(\frac{b+a}{2}\right)\cos\left(\frac{b+a }{2}\right)\bigg]}_{\text{This is }\sin(b+a)}\, \underbrace{\bigg[2\sin\left(\frac{b-a}{2}\right)\cos\left(\frac{b-a}{2}\right)\bigg]}_{\text{This is }\sin(b-a)}$

. . . . . . . $=\;4\,\sin(b+a)\,\sin(b-a)$

**pencil09** · December 31st 2009, 08:02 AM

Originally Posted by cakeboby

Prove the following identities:
$\frac {sin3a}{sina} - \frac {sin3b}{sinb}= 4sin(b+a)sin(b-a)$

Here's what I did:
L.H.S.
$=3-4sin^2a-(3-4sin^2b)$
$=4(sin^2 b - sin^2 a)$
???

Thanks in advance.

you can factorize $(sin^2 b - sin^2 a)$ to continue the equation
become
$=4(sin (b) - sin (a)) (sin (b) + sin (a))$
and continue with sum to product identities.
for help use this: http://en.wikipedia.org/wiki/List_of...uct_identities

**Soroban** · December 31st 2009, 09:12 AM

Hello, cakeboby!

Do you want to surprise/impress/terrify your professor?

Prove: . $\frac{\sin3a}{\sin a } - \frac{\in3b}{\sin b}\:=\: 4\sin(b+a)\sin(b-a)$

Here's what I did:
. . $LHS\;=\;(3-4\sin^2\!a)-(3-4\sin^2\!b) \;=\;4(\sin^2\! b - \sin^2\! a)$ ??

You have: . $4\left[\sin^2\!b-\sin^2\!a\right]$

Subtract and add $\sin^2\!a\sin^2\!b:\quad 4\left[\sin^2\!b {\color{red}\:- \:\sin^2\!a\sin^2\!b} \:- \sin^2\!a {\color{red}\:+\: \sin^2\!a\sin^2\!b}\right]$

Factor: . $4\bigg[\sin^2\!b\left(1-\sin^2\!a\right) - \sin^2\!a\left(1-\sin^2\!b\right)\bigg] \;=\; 4\bigg[\sin^2\!b\cos^2\!a - \sin^2\!a\cos^2\!b\bigg]$

Factor: . $4\bigg(\sin b\cos a + \sin a \cos b\bigg)\bigg(\sin b\cos a - \sin a\cos b\bigg) \;=\;4\sin(b+a)\sin(b-a)$

. . . ta-DAA!

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