Another identity, brain still not working

**sinewave85** · October 28th 2006, 06:44 PM

Ok, I know it seems like I am spunging, but there are thirty of these on this assignement, only stumped on two -- so far. The identity is:

(sin(x) + sin(3x))/(cos(x) + cos(3x) = tan(2x)

The only identities I have covered so far in my book are double-angle, half-angle, sum and difference, and the basics (reciprocal, quotentient, pythagorean).

I have tried subbing the tan(2x) for 2tan(x)/(1 - tan^2(x)), but I can't figure out where to go from there. I also tried to work on the other side -- sin(2x + x), use the sum identity, etc. -- but got really confused. I am doing trig on a distance learning course, and I don't really have access to a teacher or fellow students to talk it over with, so I really appreciate the help! I have been working on these way too long.

**ThePerfectHacker** · October 28th 2006, 06:51 PM

Originally Posted by sinewave85

Ok, I know it seems like I am spunging, but there are thirty of these on this assignement, only stumped on two -- so far. The identity is:

(sin(x) + sin(3x))/(cos(x) + cos(3x) = tan(2x)

The only identities I have covered so far in my book are double-angle, half-angle, sum and difference, and the basics (reciprocal, quotentient, pythagorean).

I have tried subbing the tan(2x) for 2tan(x)/(1 - tan^2(x)), but I can't figure out where to go from there. I also tried to work on the other side -- sin(2x + x), use the sum identity, etc. -- but got really confused. I am doing trig on a distance learning course, and I don't really have access to a teacher or fellow students to talk it over with, so I really appreciate the help! I have been working on these way too long.

Use,
$\sin x+\sin y=2\sin \left( \frac{x+y}{2}\right) \cos \left( \frac{x-y}{2} \right)$
Und,
$\cos x+\cos y=2\cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$
Thus,
$\sin x+\sin 3x=2\sin 2x \cos x$
$\cos x+\cos 3x=2\cos 2x \cos x$
Thus,
$\frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\frac{2\sin 3x\cos x}{2\cos 2x\cos x}=\frac{\sin 2x}{\cos 2x}=\tan 2x$

**sinewave85** · October 28th 2006, 07:36 PM

I am sorry, but I am not really familiar with that identity.
I have learned sin(x + y), cos(x + y), but not sin(x) + sin(y)
or cos(x) + cos(y). Your solution is very efficient, but I don't
think I can use it. I have reworked the rs to:

(-4sin^3(a) + 4sin(a))/(4cos^(3)a - 2cos(a))

I there anywhere to go with that?

**sinewave85** · October 28th 2006, 07:39 PM

Originally Posted by sinewave85

I have reworked the rs to:

(-4sin^3(a) + 4sin(a))/(4cos^(3)a - 2cos(a))

Correction: I have reworked the left side.

**Soroban** · October 29th 2006, 07:00 AM

Hello, sinewave85!

You know enough to derive the necessary formulas . . .
but it takes more work than we should be expected to do!

$\frac{\sin(x) + \sin(3x)}{\cos(x) + \cos(3x)} \:= \:\tan(2x)$

We need the Triple-angle Identities.
We can keep them on a handy list (I do) . . . or derive them.

$\sin(3x) \;= \;\sin(2x + x) \;=\;\sin(x)\cos(2x) + \sin(2x)\cos(x)$

. . . . . . $= \;\sin(x)\left[1 - 2\sin^2(x)\right] + \left[2\sin(x)\cos(x)\right]\cos(x)$

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

. . . . . . $= \;\sin(x) - 2\sin^3(x) + 2\sin(x)\left[1 - \sin^2(x)\right]$

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

Therefore: . $\boxed{\sin(3x) \;=\;3\sin(x) - 4\sin^3(x)}$

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

. . . . . . $= \;\cos(x)\left[2\cos^2(x) - 1\right] -\sin(x)\left[2\sin(x)\cos(x)\right]$

. . . . . . $= \;2\cos^3(x) - \cos(x) - 2\sin^2(x)\cos(x)$

. . . . . . $= \;2\cos^3(x) - \cos(x) - 2\left[1 - \cos^2(x)\right]\cos(x)$

. . . . . . $= \;2\cos^3(x) - \cos(x) - 2\cos(x) + 2\cos^3(x)$

Therefore: . $\boxed{\cos(3x)\;=\;4\cos^3(x) - 3\cos(x)}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Back to the problem . . .

We have: . $\frac{\sin x + \sin(3x)}{\cos(x) + \cos(3x)} \;= \;\frac{\sin(x) + \left[3\sin(x) - 4\sin^3(x)\right]}{\cos(x) + \left[4\cos^3(x) - 3\cos(x)\right]} \;=\;\frac{4\sin(x) - 4\sin^3(x)}{4\cos^3(x) - 2\cos(x)}$

. . . . . . $= \;\frac{4\sin(x)\left[1 - \sin^2(x)\right]}{2\cos(x)\left[2\cos^2(x) - 1\right]} \;=\;\frac{4\sin(x)\cos^2(x)}{2\cos(x)\left[2\cos^2(x) - 1\right]} \;=\;\frac{2\sin(x)\cos(x)}{2\cos^2(x) - 1}$

. . . . . . $= \;\frac{2\sin(x)\cos(x)}{2\left[1 - \sin^2(x)\right]-1} \;=\;\frac{2\sin(x)\cos(x)}{1 - 2\sin^2(x)}$

Divide top and bottom by $\cos^2(x)$

. . $\frac{\frac{2\sin(x)\cos(x)}{\cos^2(x)}}{\frac{1}{ \cos^2(x)} - \frac{2\sin^2(x)}{\cos^2{x}}} \;= \;\frac{2\frac{\sin(x)}{\cos(x)}}{\sec^2(x) - 2\left(\frac{\sin(x)}{\cos(x)}\right)^2} \;= \;\frac{2\tan(x)}{\left[\tan^2(x) + 1\right] - 2\tan^2(x)}$

. . $= \;\frac{2\tan(x)}{1 - \tan^2(x)} \;= \;\tan(2x)$ . . . There!

I need a nap . . .

**sinewave85** · October 30th 2006, 08:06 AM

Thanks for all of the help! I do appreciate all of the work you did, and I hope you got your nap.

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