Miscellaneous · Q4: $\sin^{-1}\frac{8}{17}+\sin^{-1}\frac35=\tan^{-1}\frac{77}{36}$

Basics you need

Solution

1. Name the two angles and locate them

   Let

   $$A = \sin^{-1}\frac{8}{17}, \qquad B = \sin^{-1}\frac35.$$

   Then $\sin A = \frac{8}{17}$ and $\sin B = \frac35$⁠, with $A$⁠, $B$ in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠. Both sines are positive and less than 1, so

   $$0 < A < \frac{\pi}{2}, \qquad 0 < B < \frac{\pi}{2}.$$

   We must prove $A + B = \tan^{-1}\frac{77}{36}$⁠.

   Each inverse-trig term is an angle. Naming them turns the problem into ordinary trigonometry of $A$ and $B$⁠. Knowing that both are acute fixes the signs of their cosines and tangents (all positive), and gives $0 < A+B < \pi$⁠, which Step 4 uses.
   $y = \sin^{-1}x$ means $\sin y = x$ and $y \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (Basic 1).

2. Build the two triangles

   For $A$⁠: opposite 8, hypotenuse 17, so the adjacent side is

   $$\sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15.$$

   For $B$⁠: opposite 3, hypotenuse 5, adjacent $\sqrt{25 - 9} = 4$⁠. Reading off the ratios:

   $$\begin{aligned} \cos A &= \tfrac{15}{17}, & \tan A &= \tfrac{8}{15}, \\[1mm] \cos B &= \tfrac45, & \tan B &= \tfrac34. \end{aligned}$$
   The right-hand side of the question is a $\tan^{-1}$⁠, so we want tangents. The triangles supply them. The cosines are positive because $A$ and $B$ are acute (Step 1); with $\cos^2 A = 1 - \sin^2 A$ you would otherwise have to choose the sign yourself.
   Pythagorean triples: $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ and $3^2 + 4^2 = 5^2$⁠. Lesson: the triangle trick

3. Add the angles: $\tan(A+B) = \frac{77}{36}$

   First the pieces of the addition formula:

   $$\begin{aligned} \tan A + \tan B &= \frac{8}{15} + \frac34 = \frac{32 + 45}{60} = \frac{77}{60}, \\[1mm] \tan A \tan B &= \frac{8}{15}\cdot\frac34 = \frac{24}{60} = \frac25. \end{aligned}$$

   Then

   $$\begin{aligned} \tan(A+B) &= \frac{\tan A + \tan B}{1 - \tan A\tan B} \\[1mm] &= \frac{77/60}{1 - 2/5} = \frac{77/60}{3/5} \\[1mm] &= \frac{77}{60}\cdot\frac{5}{3} = \frac{77}{36}. \end{aligned}$$

   For the next step we also note the cosine and sine of $A+B$⁠:

   $$\begin{aligned} \cos(A+B) &= \tfrac{15}{17}\cdot\tfrac45 - \tfrac{8}{17}\cdot\tfrac35 \\[1mm] &= \tfrac{60 - 24}{85} = \tfrac{36}{85}, \\[2mm] \sin(A+B) &= \tfrac{8}{17}\cdot\tfrac45 + \tfrac{15}{17}\cdot\tfrac35 \\[1mm] &= \tfrac{32 + 45}{85} = \tfrac{77}{85}. \end{aligned}$$

   So $A+B$ belongs to a 36, 77, 85 triangle (check: $36^2 + 77^2 = 1296 + 5929 = 7225 = 85^2$⁠).

   The denominator $1 - \tan A\tan B = \frac35$ is not zero, so $\tan(A+B)$ is defined and the formula applies. Its sign matters too: a positive denominator with a positive numerator gives a positive tangent, which Step 4 uses.
   $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$⁠, $\cos(A+B) = \cos A\cos B - \sin A\sin B$⁠, $\sin(A+B) = \sin A\cos B + \cos A\sin B$ (Class XI, Chapter 3).

4. The crucial check: $A+B$ is acute

   Step 3 shows only that $A+B$ and $\tan^{-1}\frac{77}{36}$ have the same tangent. They are the same angle only if $A+B$ lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠.

   From Step 1, adding the two inequalities,

   $$0 < A + B < \pi.$$

   On this interval, the angle $\frac{\pi}{2}$ is excluded because $\tan(A+B)$ is defined, and the angles in $\left(\frac{\pi}{2},\pi\right)$ are excluded because their tangent is negative while $\tan(A+B) = \frac{77}{36} > 0$⁠. Hence

   $$0 < A + B < \frac{\pi}{2}.$$

   The cosine agrees: $\cos(A+B) = \frac{36}{85} > 0$⁠, which is impossible for an angle in $\left[\frac{\pi}{2},\pi\right)$⁠.

   Tangent repeats every $\pi$⁠, so "equal tangents" leaves open $A+B = \tan^{-1}\frac{77}{36} + k\pi$ for some integer $k$ (positive, negative or zero). Only $k = 0$ gives an angle in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠. The acute check is exactly what pins down $k = 0$⁠. In general, $A+B$ (with $A$⁠, $B$ acute) is acute exactly when $\tan A\tan B < 1$⁠; here $\tan A\tan B = \frac25$⁠.
   Signs: on $\left(\frac{\pi}{2},\pi\right)$ tangent and cosine are negative (Basic 4). For a case where the check fails, see Miscellaneous Q3, in its final step.

5. Conclude

   $A+B$ lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ (Step 4) and $\tan(A+B) = \frac{77}{36}$ (Step 3). Tangent takes each value only once on that interval, so

   $$\begin{aligned} \tan^{-1}\frac{77}{36} &= \tan^{-1}\bigl(\tan(A+B)\bigr) = A + B \\[1mm] &= \sin^{-1}\frac{8}{17} + \sin^{-1}\frac35. \end{aligned}$$

   This is the required result.

   The chain is complete: named angles (Step 1), correct ratios (Step 2), the tangent of the sum (Step 3), and the sum placed in the range where $\tan^{-1}$ undoes $\tan$ (Step 4).

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