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).