Knowing that derivative of sinx is cos⁡x and derivative of cosx is -senx, prove by induction, that the 2n derivative of sinx = ((- 1) ^ n ) * sin⁡x⁡ and the 2n derivative of cos⁡x = ((- 1) ^ n)* cos⁡x.

so, check n=1. The 2nd derivative of sinx is -sinx since

d/dx(sinx) = cosx
d/dx(cosx) = -sinx
Now assume it is true for k=2n. What about 2n+2?
y(2n)sinx = (-1)n sinx
y(2n+1)sinx = (-1)n cosx
y(2n+2)sinx = -(-1)n cosx = (-1)n+1 sinx
QED