Prove: sin3x-sinx/cos3x-cosx=-cot2

I think I would use the sum formulas, changing sin3x to sin(2x+x) and cos3x to cos(2x+x), then gather terms.

To prove the given identity: (sin3x - sinx) / (cos3x - cosx) = -cot2x, we can manipulate the left side of the equation to obtain the right side.

1) Start with the left side of the equation: (sin3x - sinx) / (cos3x - cosx)

2) Apply the trigonometric identity for the difference of sines: sin(A) - sin(B) = 2sin((A-B)/2)cos((A+B)/2). We get:

(2sin((3x-x)/2)cos((3x+x)/2)) / (cos3x - cosx)

Simplifying further, we have:

(2sin(2x)cos(2x)) / (cos3x - cosx)

3) Now, apply the trigonometric identity for the difference of cosines: cos(A) - cos(B) = -2sin((A+B)/2)sin((A-B)/2).

In our equation, this means we can rewrite the denominator as:

-2sin((3x+x)/2)sin((3x-x)/2)

Simplifying further, we have:

-2sin(2x)sin(x)

4) Substituting this back into our original equation, we get:

(2sin(2x)cos(2x)) / (-2sin(2x)sin(x))

5) Now, cancel out the common factors of 2 and sin(2x):

(cos(2x) / -sin(x))

6) Applying the identity cot(x) = cos(x) / sin(x), we have:

-cot(2x)

Therefore, we have proved that (sin3x - sinx) / (cos3x - cosx) = -cot(2x).