# Thread: differentiation of inverse trignometric equation continuation

2. ## Re: differentiation of inverse trignometric equation continuation

You ignored the absolute value signs. Don't ignore them and you will get the two answers you show. They are each correct on different intervals. Find the intervals that make each true.

3. ## Re: differentiation of inverse trignometric equation continuation

In the book they have given it as x/2 and on differentiation, it is 1/2. But, my friend proceeded by rationalizing the denominator by multiplying the conjugate. I could not understand which method I have to adopt and why? it was given only that x lies between 0 and Pi/2. That means x is in the first quadrant.

4. ## Re: differentiation of inverse trignometric equation continuation

$\sin \dfrac{x}{2} + \cos \dfrac{x}{2}\ge 0$ when $4n\pi -\dfrac{\pi}{2} \le x \le \dfrac{3\pi}{2}, n\in \mathbb{N}$ (this covers quadrant 1 when $n=0$).

$\sin \dfrac{x}{2}-\cos \dfrac{x}{2} < 0$ when $4n\pi - \dfrac{3\pi}{2} < x < 4n\pi + \dfrac{\pi}{2}, n \in \mathbb{N}$ (this covers quadrant 1 when $n=0$).

So, $\left| \sin \dfrac{x}{2} + \cos \dfrac{x}{2} \right| = \sin \dfrac{x}{2} + \cos \dfrac{x}{2}$ in quadrant 1 while $\left| \sin \dfrac{x}{2} - \cos \dfrac{x}{2} \right| = \cos \dfrac{x}{2} - \sin \dfrac{x}{2}$ when x is in quadrant 1.

Then, you have:

$\displaystyle \dfrac{ \left| \sin \dfrac{x}{2} + \cos \dfrac{x}{2} \right| + \left| \sin \dfrac{x}{2} - \cos \dfrac{x}{2} \right| }{ \left| \sin \dfrac{x}{2} + \cos \dfrac{x}{2} \right| - \left| \sin \dfrac{x}{2} - \cos \dfrac{x}{2} \right| } = \dfrac{2\cos \dfrac{x}{2} }{2\sin \dfrac{x}{2}}$ when x is in quadrant 1.

Thus, you have $\cot^{-1}\cot \dfrac{x}{2} = \dfrac{x}{2}$ when x is in quadrant 1. Your method works just fine. You just needed to check the intervals where each answer was true.

5. ## Re: differentiation of inverse trignometric equation continuation

Excellent Sir. Your reply was very useful.