How is the Conjugate of Real Trigonometric Functions Handled?

[kolycholy]

okay, so this particular equation involves me writing conjugate of either sin or cos, but hows that possible considering they both are real in the given problem?

maybe i should convert sin and cos into their exponential form first?

but then wt would be the conjugate of this expression-----> e^2j +e^(-2j)?

 

[jpr0]

okay, so this particular equation involves me writing conjugate of either sin or cos, but hows that possible considering they both are real in the given problem?

maybe i should convert sin and cos into their exponential form first?

but then wt would be the conjugate of this expression-----> e^2j +e^(-2j)?

If you have to take the complex conjugate of a real quantity, say $z$, then $z$ is its own complex conjugate, i.e. $z=z^{\ast}$. This follows from the fact that the real part of a complex number and the real part of its conjugate are always the same by definition:

$$z = x + iy$$
$$z^{\ast} = x - iy$$

where $x\,,y$ are real.