A question about trigonometric identities

[acen_gr]

Is this identity possible?

cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}

Thanks!

 

[Mark44]

Is this homework?

 

[acen_gr]

Is this homework?

Not homework. I'm just asking if it's possible for the expressions on both sides to be identical or the equation is an identity or not. Because I've tried to work on it but I couldn't make them identical.

 

[HallsofIvy]

One obvious point is that if x= 0, the left side is 1 but the right side does not exist. And if you don't like "does not exist", try evaluating both sides at x= 0.01.

 

[uart]

Is this identity possible?

cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}

Thanks!

Try cross multiplying to get:

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

and then apply the half sum/difference identities to all of the sin cos products on each side of the equation. It comes out fairly easy with this step.

Use:
\sin a \cos b = \frac{1}{2} \left[ \sin(a+b) + \sin(a-b) \right]

 

[mathman]

An alternate approach is to convert to a polynomial equation in u = e^ix.
For example: cot(2x) = i(u²+u^-2)/(u²-u^-2) = i(u⁴+1)/(u⁴-1).
Do the similar steps for the right hand side, clear the denominators and end up with an identity.

 

[acen_gr]

Try cross multiplying to get:

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

and then apply the half sum/difference identities to all of the sin cos products on each side of the equation. It comes out fairly easy with this step.

Use:
\sin a \cos b = \frac{1}{2} \left[ \sin(a+b) + \sin(a-b) \right]

\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)
\cos 2x \sin3x + \cos2x sinx = \sin2x \cos3x + \sin2x cosx

Should I end here? I think both doesn't equal up. Or should I go further by extracting cos2x, sin2x, cos3x, and sin3x ?

 

[mathman]

\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)
\cos 2x \sin3x + \cos2x sinx = \sin2x \cos3x + \sin2x cosx

Should I end here? I think both doesn't equal up. Or should I go further by extracting cos2x, sin2x, cos3x, and sin3x ?

You need to end up with sinx and cosx only, using identities for cos2x, etc. I think my method (using representation in e^ix, etc.) might be easier.

 

[uart]

\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)
\cos 2x \sin3x + \cos2x sinx = \sin2x \cos3x + \sin2x cosx

Should I end here?

No, apply the half sum-difference formula (that I gave above) to each of the sin-cos products.