If p=cis theta and q= cis phi, show that

p-q/p+q=i tan ((theta-phi)/2)

To demonstrate that $\frac{p-q}{p+q}=i\tan \left(\frac{\theta -\varphi }{2}\right)$ given that $p=\text{cis}(\theta )$ and $q=\text{cis}(\varphi )$, we first need to understand what the notation $\text{cis}(\theta )$ means. The term $\text{cis}(\theta )$ is shorthand for $\cos (\theta )+i\sin (\theta )$. This representation allows us to work with complex numbers in a more manageable way.

Step-by-Step Breakdown

Let's start by substituting the definitions of $p$ and $q$ into our expression:

• Substituting $p$ and $q$:

We have:

p = \cos(\theta) + i\sin(\theta)

q = \cos(\phi) + i\sin(\phi)

Now, we can compute $p-q$ and $p+q$:

• Calculating $p-q$:

p - q = (\cos(\theta) - \cos(\phi)) + i(\sin(\theta) - \sin(\phi))

• Calculating $p+q$:

p + q = (\cos(\theta) + \cos(\phi)) + i(\sin(\theta) + \sin(\phi))

Forming the Quotient

Next, we need to find the quotient $\frac{p-q}{p+q}$:

\frac{p - q}{p + q} = \frac{(\cos(\theta) - \cos(\phi)) + i(\sin(\theta) - \sin(\phi)}{(\cos(\theta) + \cos(\phi)) + i(\sin(\theta) + \sin(\phi))}

Using Trigonometric Identities

To simplify this expression, we can use the trigonometric identities for the difference and sum of angles:

• Difference of Cosines: $\cos (\theta )-\cos (\varphi )=-2\sin \left(\frac{\theta +\varphi }{2}\right)\sin \left(\frac{\theta -\varphi }{2}\right)$
• Difference of Sines: $\sin (\theta )-\sin (\varphi )=2\cos \left(\frac{\theta +\varphi }{2}\right)\sin \left(\frac{\theta -\varphi }{2}\right)$
• Sum of Cosines: $\cos (\theta )+\cos (\varphi )=2\cos \left(\frac{\theta +\varphi }{2}\right)\cos \left(\frac{\theta -\varphi }{2}\right)$
• Sum of Sines: $\sin (\theta )+\sin (\varphi )=2\sin \left(\frac{\theta +\varphi }{2}\right)\cos \left(\frac{\theta -\varphi }{2}\right)$

Substituting Back

Now, substituting these identities back into our quotient:

\frac{p - q}{p + q} = \frac{-2\sin\left(\frac{\theta + \phi}{2}\right)\sin\left(\frac{\theta - \phi}{2}\right) + i \cdot 2\cos\left(\frac{\theta + \phi}{2}\right)\sin\left(\frac{\theta - \phi}{2}\right)}{2\cos\left(\frac{\theta + \phi}{2}\right)\cos\left(\frac{\theta - \phi}{2}\right) + i \cdot 2\sin\left(\frac{\theta + \phi}{2}\right)\cos\left(\frac{\theta - \phi}{2}\right)}

We can simplify this by canceling the common factor of 2:

\frac{-\sin\left(\frac{\theta + \phi}{2}\right)\sin\left(\frac{\theta - \phi}{2}\right) + i \cdot \cos\left(\frac{\theta + \phi}{2}\right)\sin\left(\frac{\theta - \phi}{2}\right)}{\cos\left(\frac{\theta + \phi}{2}\right)\cos\left(\frac{\theta - \phi}{2}\right) + i \cdot \sin\left(\frac{\theta + \phi}{2}\right)\cos\left(\frac{\theta - \phi}{2}\right)}

Final Simplification

Now, we can factor out $\sin \left(\frac{\theta -\varphi }{2}\right)$ from the numerator:

\frac{\sin\left(\frac{\theta - \phi}{2}\right)\left(-\sin\left(\frac{\theta + \phi}{2}\right) + i \cdot \cos\left(\frac{\theta + \phi}{2}\right)\right)}{\cos\left(\frac{\theta - \phi}{2}\right)\left(\cos\left(\frac{\theta + \phi}{2}\right) + i \cdot \sin\left(\frac{\theta + \phi}{2}\right)\right)}

Recognizing that the expression in the numerator can be rewritten as $i$ times the tangent function, we arrive at:

\frac{p - q}{p + q} = i \tan\left(\frac{\theta - \phi}{2}\right)

Conclusion

This completes the proof that $\frac{p-q}{p+q}=i\tan \left(\frac{\theta -\varphi }{2}\right)$. By utilizing trigonometric identities and simplifying the complex expressions, we can see how the relationship holds true. This approach not only illustrates the power of complex numbers but also highlights the beauty of trigonometric relationships.