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Solve sin6x - inn4x = sin5x - sin3x
By Saif on Tuesday, November 23, 1999 - 06:38 pm:

I would be grateful if somebody can answer the question below.

Find in radians in terms of pi, the general solution of the equation.
Sin6x - Sin4x = Sin5x - Sin3x

Saif

By James Lingard (Jchl2) on Tuesday, November 23, 1999 - 09:15 pm:

Hi there.

The key to solving this question is knowing the formula for expressing sin A - sin B as a product of sines and cosines. Rather than just quote this result at you, I'll give a proof.

I'll assume that you're familiar with (or at least have come across) the formulae for cos(p ± q) and sin(p ± q), namely:

sin(p + q) = sin p cos q - cos p sin q
sin(p - q) = sin p cos q + cos p sin q

cos(p + q) = cos p cos q - sin p sin q
cos(p - q) = cos p cos q + sin p sin q

If you haven't seen these before then write back and I'll give a proof.

Now, subtracting the second equation from the first gives:

sin(p + q) - sin(p - q) = -2 cos p sin q (*)

since the other two terms on the right hand side cancel each other. What we're looking for is similar to this, but with A instead of p + q and B instead of p - q. So do the following:

Let

A = p + q
B = p - q

Then

A + B = 2p (adding the two equations)
A - B = 2q (subtracting the two equations)

And so

p = (A + B)/2
q = (A - B)/2

Substituting these into the equation marked (*), we get:

sin A - sin B = -2 cos((A + B)/2) sin((A - B)/2)

(There are similar equations to this one for sin A + sin B, and for cos A ± cos B. They can be derived in a very similar way to what I've done here - see if you can do that for yourself. Hint: one of the other equations will come in handy later in the solution of the question.)

Now you can use this equation to simplify the question, and with my hint above see if you can do the rest yourself. If any of what I've done here isn't clear, or if you need some more help with the question, then please let me know.

James.