Miscellaneous · Q3: $2\sin^{-1}\frac35=\tan^{-1}\frac{24}{7}$

Basics you need

Solution

1. Name the angle and locate it

   Let

   $$\theta = \sin^{-1}\frac35.$$

   By definition of the principal value, $\sin\theta = \frac35$ and $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$⁠. Sine is $\le 0$ on $\left[-\frac{\pi}{2}, 0\right]$⁠, so $\sin\theta > 0$ means $\theta$ is not negative and not zero; since $\sin\theta \ne 1$⁠, $\theta \ne \frac{\pi}{2}$⁠. So

   $$0 < \theta < \frac{\pi}{2}.$$

   What we must prove becomes $2\theta = \tan^{-1}\frac{24}{7}$⁠.

   An inverse function value is just an angle. Naming it lets us use ordinary trigonometry on it. Locating it (first quadrant) matters at once: it fixes the sign of $\cos\theta$ in the next step, and later it is the starting point for locating $2\theta$⁠.
   $y = \sin^{-1}x$ means $\sin y = x$ and $y \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (Basic 1).

2. Build the right triangle: $\cos\theta = \frac45$⁠, $\tan\theta = \frac34$

   Because $\theta$ is acute, it is an angle of a right triangle with opposite side 3 and hypotenuse 5. By Pythagoras the adjacent side is

   $$\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4.$$

   Reading off the ratios,

   $$\cos\theta = \frac45, \qquad \tan\theta = \frac{3}{4}.$$
   The same result follows from $\sin^2\theta + \cos^2\theta = 1$⁠: $\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$⁠, so $\cos\theta = \pm\frac45$⁠. The positive sign is forced by Step 1: on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ cosine is never negative. The triangle picture builds that sign in automatically, which is why it only works for acute angles.
   Pythagoras: (hypotenuse)$^2$ = (opposite)$^2$ + (adjacent)$^2$⁠. Here $3^2 + 4^2 = 25 = 5^2$⁠. Lesson: the triangle trick

3. Double the angle: $\tan 2\theta = \frac{24}{7}$

   Use the double-angle formula with $\tan\theta = \frac34$⁠. Here $\tan^2\theta = \frac{9}{16} \ne 1$⁠, so the formula applies:

   $$\begin{aligned} \tan 2\theta &= \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2\cdot\frac34}{1 - \frac{9}{16}} \\[1mm] &= \frac{3/2}{7/16} = \frac32\cdot\frac{16}{7} = \frac{24}{7}. \end{aligned}$$

   For later use, the sine and cosine of $2\theta$ are

   $$\begin{aligned} \sin 2\theta &= 2\sin\theta\cos\theta \\[1mm] &= 2\cdot\tfrac35\cdot\tfrac45 = \tfrac{24}{25}, \\[2mm] \cos 2\theta &= \cos^2\theta - \sin^2\theta \\[1mm] &= \tfrac{16 - 9}{25} = \tfrac{7}{25}. \end{aligned}$$

   So $2\theta$ belongs to a 7, 24, 25 triangle (check: $7^2 + 24^2 = 49 + 576 = 625 = 25^2$⁠).

   The right-hand side of the question is a $\tan^{-1}$⁠, so we convert the left-hand side into a tangent. The formula for $\tan 2\theta$ needs only $\tan\theta$⁠, which the triangle supplied.
   $\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}$⁠, $\sin 2\theta = 2\sin\theta\cos\theta$⁠, $\cos 2\theta = \cos^2\theta - \sin^2\theta$ (Class XI, Chapter 3).

4. The crucial check: $2\theta$ lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

   From Step 3, $\tan 2\theta = \frac{24}{7}$⁠. But $\tan^{-1}\frac{24}{7}$ is the angle in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ with that tangent, so we must show $2\theta$ is in that interval.

   Compare $\theta$ with $\frac{\pi}{4}$⁠. We have $\sin\frac{\pi}{4} = \frac{1}{\sqrt2}$⁠. Both $\frac35$ and $\frac{1}{\sqrt2}$ are positive, so compare their squares:

   $$\left(\frac35\right)^2 = \frac{9}{25} = 0.36 < 0.5 = \left(\frac{1}{\sqrt2}\right)^2.$$

   So $\sin\theta < \sin\frac{\pi}{4}$⁠. Sine is increasing on $\left[0,\frac{\pi}{2}\right]$⁠, and both angles lie there, so

   $$0 < \theta < \frac{\pi}{4}, \qquad\text{hence}\qquad 0 < 2\theta < \frac{\pi}{2}.$$

   A second way to see it: $\cos 2\theta = \frac{7}{25} > 0$ and $\sin 2\theta = \frac{24}{25} > 0$⁠. Since $0 < 2\theta < \pi$ and its cosine is positive, $2\theta$ is in the first quadrant.

   Tangent repeats every $\pi$⁠. The equation $\tan 2\theta = \frac{24}{7}$ is also true of $2\theta \pm \pi$⁠, $2\theta \pm 2\pi$⁠, and so on. Only one of these angles lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠, and only that one equals $\tan^{-1}\frac{24}{7}$⁠. Without this step the proof would have shown only that the two sides have the same tangent, not that they are equal.
   $\tan^{-1}(\tan x) = x$ only for $x \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ (Basic 4). Sine increasing on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (Basic 5).

5. Conclude

   $2\theta$ lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ (Step 4) and $\tan 2\theta = \frac{24}{7}$ (Step 3). Tangent takes each value only once on that interval, so $2\theta$ is the principal value:

   $$\tan^{-1}\frac{24}{7} = \tan^{-1}(\tan 2\theta) = 2\theta = 2\sin^{-1}\frac35.$$

   This is the required result.

   Every link is now justified: Step 1 defined $\theta$⁠, Step 2 found its ratios with the correct signs, Step 3 found $\tan 2\theta$⁠, and Step 4 put $2\theta$ in the one interval where $\tan^{-1}$ undoes $\tan$⁠.

6. Why the range check is not a formality

   Try the same method on $\varphi = \sin^{-1}\frac45$⁠. The 3, 4, 5 triangle now gives $\tan\varphi = \frac43$⁠, and

   $$\tan 2\varphi = \frac{2\cdot\frac43}{1 - \frac{16}{9}} = \frac{\frac83}{-\frac79} = -\frac{24}{7}.$$

   But $\frac45 > \frac{1}{\sqrt2}$⁠, so $\varphi > \frac{\pi}{4}$ and $2\varphi > \frac{\pi}{2}$ (in fact $2\varphi \approx 1.8546$⁠). This angle is outside the range of $\tan^{-1}$⁠, so $2\sin^{-1}\frac45 \ne \tan^{-1}\left(-\frac{24}{7}\right)$⁠, which is negative (about $-1.2870$⁠). The correct statement is

   $$2\sin^{-1}\frac45 = \pi - \tan^{-1}\frac{24}{7}.$$
   The algebra in Step 3 is identical in both cases; only the range check tells them apart. That is why Step 4 is the heart of the proof, not an afterthought.

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