Exercise 2.2 · Q9: $\tan\frac12\Big[\sin^{-1}\frac{2x}{1+x^2}$ $+\cos^{-1}\frac{1-y^2}{1+y^2}\Big]$

Basics you need

Solution

1. $\sin2\phi$ and $\cos2\phi$ in terms of $\tan\phi$

   For any $\phi$ with $\cos\phi\ne0$⁠, we have $\cos^2\phi=\dfrac{1}{\sec^2\phi}=\dfrac{1}{1+\tan^2\phi}$⁠. Hence

   $$\begin{aligned} \sin2\phi &= 2\sin\phi\cos\phi \\ &= 2\tan\phi\cos^2\phi \\ &= \frac{2\tan\phi}{1+\tan^2\phi}, \end{aligned}$$ $$\begin{aligned} \cos2\phi &= \cos^2\phi-\sin^2\phi \\ &= \cos^2\phi\left(1-\tan^2\phi\right) \\ &= \frac{1-\tan^2\phi}{1+\tan^2\phi}. \end{aligned}$$
   These are exactly the two fractions in the question, with $x$ or $y$ in place of $\tan\phi$⁠. The rewriting uses $\sin\phi=\tan\phi\cos\phi$ and then $\cos^2\phi=\frac{1}{1+\tan^2\phi}$⁠.
   $1+\tan^2\phi=\sec^2\phi$ comes from dividing $\sin^2\phi+\cos^2\phi=1$ by $\cos^2\phi$⁠.

2. Set $x=\tan\alpha$⁠, $y=\tan\beta$ and find the ranges

   Let $\alpha=\tan^{-1}x$ and $\beta=\tan^{-1}y$⁠, so $x=\tan\alpha$ and $y=\tan\beta$⁠, with both angles in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠. Because $\tan^{-1}$ is increasing and $\tan^{-1}(\pm1)=\pm\frac{\pi}{4}$⁠, $\tan^{-1}0=0$⁠:

   • $-1<x<1$ gives $-\frac{\pi}{4}<\alpha<\frac{\pi}{4}$⁠, so $-\frac{\pi}{2}<2\alpha<\frac{\pi}{2}$.
   • $y>0$ gives $0<\beta<\frac{\pi}{2}$⁠, so $0<2\beta<\pi$.
   The substitutions are always allowed, since $\tan^{-1}$ accepts every real number. What matters is where the angles land. The two highlighted facts are exactly what Steps 3 and 4 need.
   $\tan\frac{\pi}{4}=1$⁠. An increasing function preserves inequalities. Basics: special angles

3. The $\sin^{-1}$ term is $2\alpha$

   By Step 1 with $\phi=\alpha$⁠,

   $$\frac{2x}{1+x^2}=\frac{2\tan\alpha}{1+\tan^2\alpha}=\sin2\alpha.$$

   So $\sin^{-1}\dfrac{2x}{1+x^2}=\sin^{-1}(\sin2\alpha)=2\alpha$⁠, because $2\alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ (Step 2) lies inside the principal range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ of $\sin^{-1}$⁠.

   This is where $|x|<1$ is used. If $|x|>1$⁠, then $2\alpha$ would be outside $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and $\sin^{-1}(\sin2\alpha)$ would not be $2\alpha$⁠. For example, $x=2$ gives $\sin^{-1}\frac45=\pi-2\tan^{-1}2$⁠, not $2\tan^{-1}2$⁠.
   $\sin^{-1}(\sin u)=u$ for $u\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (Section 2.3). Lesson: the folding rules

4. The $\cos^{-1}$ term is $2\beta$

   By Step 1 with $\phi=\beta$⁠,

   $$\frac{1-y^2}{1+y^2}=\frac{1-\tan^2\beta}{1+\tan^2\beta}=\cos2\beta.$$

   So $\cos^{-1}\dfrac{1-y^2}{1+y^2}=\cos^{-1}(\cos2\beta)$⁠, and this is $2\beta$⁠, because $2\beta\in(0,\pi)$ (Step 2) lies inside the principal range $[0,\pi]$ of $\cos^{-1}$⁠.

   This is where $y>0$ is used. If $y<0$⁠, then $2\beta\in(-\pi,0)$ and $\cos^{-1}(\cos2\beta)=-2\beta$⁠, which changes the final answer to $\frac{x-y}{1+xy}$⁠. For example, $x=0.5$⁠, $y=-0.5$ gives the value $\frac{1}{0.75}=\frac43$⁠, while $\frac{x+y}{1-xy}=0$⁠.
   $\cos^{-1}(\cos u)=u$ for $u\in[0,\pi]$ (Section 2.3, "similar results"). Lesson: the folding rules

5. Halve the sum and use $\tan(\alpha+\beta)$

   By Steps 3 and 4,

   $$\begin{aligned} &\frac12\left[\sin^{-1}\frac{2x}{1+x^2}+\cos^{-1}\frac{1-y^2}{1+y^2}\right] \\[1mm] &\quad=\frac12(2\alpha+2\beta)=\alpha+\beta. \end{aligned}$$

   Before taking its tangent, check $\tan(\alpha+\beta)$ is defined, that is $\cos(\alpha+\beta)\ne0$⁠. Factor out $\cos\alpha\cos\beta$⁠:

   $$\begin{aligned} \cos(\alpha+\beta) &= \cos\alpha\cos\beta-\sin\alpha\sin\beta \\ &= \cos\alpha\cos\beta\,(1-\tan\alpha\tan\beta) \\ &= \cos\alpha\cos\beta\,(1-xy). \end{aligned}$$

   Here $\cos\alpha>0$ and $\cos\beta>0$ (both angles are in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠) and $1-xy>0$ (given $xy<1$⁠). So $\cos(\alpha+\beta)>0$⁠, and the addition formula applies:

   $$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{x+y}{1-xy}.$$
   This is where $xy<1$ is used: it makes the denominator $1-xy$ non-zero, so the answer exists. (At $xy=1$⁠, for instance $x=0.5$⁠, $y=2$⁠, we get $\alpha+\beta=\frac{\pi}{2}$ and the tangent is undefined.) In fact $\cos(\alpha+\beta)>0$ shows more: $\alpha+\beta$ lies in $\left(-\frac{\pi}{4},\frac{\pi}{2}\right)$⁠, so the bracket is also equal to $2\tan^{-1}\frac{x+y}{1-xy}$⁠.

   Is $xy<1$ more than we need? Yes. If $|x|<1$⁠, $y>0$ but $xy>1$ (for example $x=0.9$⁠, $y=2$⁠), Steps 2 to 4 are unchanged and $\alpha+\beta$ lies in $\left(\frac{\pi}{2},\frac{3\pi}{4}\right)$⁠, where tangent is still defined, so the addition formula still gives $\frac{x+y}{1-xy}$ (here $\frac{2.9}{-0.8}=-3.625$⁠, a negative value, as it should be in that interval). Only $xy=1$ has to be excluded. What $xy>1$ would break is the different statement that $\alpha+\beta$ equals $\tan^{-1}\frac{x+y}{1-xy}$⁠, since $\tan^{-1}$ cannot return an angle above $\frac{\pi}{2}$⁠.

   $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ (Class XI, Chapter 3). Dividing $\cos A\cos B-\sin A\sin B$ by $\cos A\cos B$ is the same algebra that proves it.

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