just wondering

[hazy shuttle]

No, it's correct.
Recall that e^(ix) = cos(x) + i sin(x). Then e^(-ix) = cos(x) - i sin(x). And you can get sin(x) and cos(x) from those.

[dull timber]

Ty

[prisma wyvern]

@vital parrot has given 1 rep to @hazy shuttle

[dull timber]

Do u know how they did that in details if so plz let me know

[tough coyote]

PDF File (apex predator)

[vocal dock]

similar thing for sinx it's just you subtract the 2nd line from the 1st and divide by 2i instead of just 2

@dull timber

[modern minnow]

The imaginary parts of e^(i x) and e^(-i x) cancel out. Remember that e^(i x) is x radians counterclockwise along the circle, so e^(-i x) is x radians clockwise; this means that the two will be opposite each other on the x-axis

like, here's e^i and e^(-i)