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.

 

[ravenprp]

The conjugate of a complex number is formed by changing the sign of the imaginary part. In the case of trigonometric functions, the imaginary part is zero, so the conjugate of either sin or cos would be the same function. In other words, the conjugate of sin or cos is just itself.

As for converting them into exponential form, that is not necessary in this case. The conjugate of e^2j + e^(-2j) would simply be e^2j - e^(-2j), as the imaginary parts have opposite signs. However, since sin and cos do not have imaginary parts, their conjugates remain unchanged.