Miscellaneous · Q6: $\cos^{-1}\frac{12}{13}+\sin^{-1}\frac35=\sin^{-1}\frac{56}{65}$

Basics you need

Solution

1. Name the two angles and locate them

   Let

   $$A = \cos^{-1}\frac{12}{13}, \qquad B = \sin^{-1}\frac35.$$

   Then $\cos A = \frac{12}{13}$ with $A \in [0,\pi]$⁠; the cosine is positive and less than 1, so $0 < A < \frac{\pi}{2}$⁠. Also $\sin B = \frac35$ with $B \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠; the sine is positive and less than 1, so $0 < B < \frac{\pi}{2}$⁠.

   We must prove $A + B = \sin^{-1}\frac{56}{65}$⁠.

   The two terms come from different inverse functions with different ranges, so each is located using its own range. Both turn out to be acute, which lets us draw them in right triangles.
   $y = \cos^{-1}x$ means $\cos y = x$⁠, $y \in [0,\pi]$⁠; $y = \sin^{-1}x$ means $\sin y = x$⁠, $y \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (Basic 1).

2. Build the two triangles

   For $A$⁠: adjacent 12, hypotenuse 13, so the opposite side is $\sqrt{169 - 144} = 5$⁠. For $B$⁠: opposite 3, hypotenuse 5, so the adjacent side is $\sqrt{25 - 9} = 4$⁠. Hence

   $$\begin{aligned} \sin A &= \tfrac{5}{13}, & \cos A &= \tfrac{12}{13}, \\[1mm] \sin B &= \tfrac35, & \cos B &= \tfrac45. \end{aligned}$$
   All four values are positive because both angles are acute (Step 1). With the identity $\sin^2 + \cos^2 = 1$ alone you would get $\pm$ signs; the location of each angle chooses the sign.
   $5^2 + 12^2 = 13^2$ and $3^2 + 4^2 = 5^2$⁠. Lesson: the triangle trick

3. Add the angles: $\sin(A+B) = \frac{56}{65}$

   $$\begin{aligned} \sin(A+B) &= \sin A\cos B + \cos A\sin B \\[1mm] &= \frac{5}{13}\cdot\frac45 + \frac{12}{13}\cdot\frac35 \\[1mm] &= \frac{20}{65} + \frac{36}{65} = \frac{56}{65}. \end{aligned}$$
   The right-hand side of the question is a $\sin^{-1}$⁠, so we compute the sine of the left-hand side. The formula needs both sines and both cosines, which is why Step 2 found all four.
   $\sin(A+B) = \sin A\cos B + \cos A\sin B$ (Class XI, Chapter 3).

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

   We know $\sin(A+B) = \frac{56}{65}$ and $\sin^{-1}\frac{56}{65}$ is the angle in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ with that sine. So we must show $A+B$ lies in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠.

   From Step 1, $0 < A + B < \pi$⁠. That is not enough: on $(0,\pi)$ both $A+B$ and $\pi - (A+B)$ have sine $\frac{56}{65}$ (Basic 4). To decide, compute the cosine:

   $$\begin{aligned} \cos(A+B) &= \cos A\cos B - \sin A\sin B \\[1mm] &= \frac{12}{13}\cdot\frac45 - \frac{5}{13}\cdot\frac35 \\[1mm] &= \frac{48}{65} - \frac{15}{65} = \frac{33}{65} > 0. \end{aligned}$$

   An angle in $\left[\frac{\pi}{2},\pi\right)$ has cosine $\le 0$⁠. Since $\cos(A+B) > 0$⁠, $A+B$ is not in that interval, so

   $$0 < A + B < \frac{\pi}{2}.$$
   This is the step that makes the proof valid. Sine alone cannot distinguish an acute angle from its obtuse supplement: in the figure, reflecting $P$ in the vertical axis gives a point with the same height at angle $\pi - (A+B)$⁠. Only the sign of the cosine (the $x$-coordinate) tells us which one $A+B$ is.
   On $\left[\frac{\pi}{2},\pi\right]$⁠, cosine is $\le 0$⁠. Basics: signs on the unit circle

5. Conclude

   $A+B$ lies in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ (Step 4) and $\sin(A+B) = \frac{56}{65}$ (Step 3). Sine takes each value only once on that interval, so

   $$\begin{aligned} \sin^{-1}\frac{56}{65} &= \sin^{-1}\bigl(\sin(A+B)\bigr) = A + B \\[1mm] &= \cos^{-1}\frac{12}{13} + \sin^{-1}\frac35. \end{aligned}$$

   This is the required result.

   Compare Miscellaneous Q5: there $\cos^{-1}\frac45$ is the same angle as our $B = \sin^{-1}\frac35$⁠, and $\cos^{-1}\frac{12}{13}$ is our $A$⁠, so both questions add the same two angles. Q5 records the cosine of the sum ($\frac{33}{65}$⁠), and this question records its sine ($\frac{56}{65}$⁠).

6. Why the cosine check is not a formality

   Take the complementary angles instead: $A' = \cos^{-1}\frac{5}{13}$ and $B' = \sin^{-1}\frac45$⁠. The triangles are the same, with the roles of the legs swapped, so $\sin A' = \frac{12}{13}$⁠, $\cos A' = \frac{5}{13}$⁠, $\sin B' = \frac45$⁠, $\cos B' = \frac35$⁠. Then

   $$\begin{aligned} \sin(A'+B') &= \tfrac{12}{13}\cdot\tfrac35 + \tfrac{5}{13}\cdot\tfrac45 \\[1mm] &= \tfrac{36}{65} + \tfrac{20}{65} = \tfrac{56}{65}, \\[2mm] \cos(A'+B') &= \tfrac{15}{65} - \tfrac{48}{65} = -\tfrac{33}{65} < 0. \end{aligned}$$

   The sine is again $\frac{56}{65}$⁠, but now $A' + B'$ is obtuse (about $2.1033$⁠, above $\frac{\pi}{2}$⁠). So $A' + B' = \pi - \sin^{-1}\frac{56}{65}$⁠, not $\sin^{-1}\frac{56}{65}$⁠.

   Same sine, different angles. The sign of the cosine is what separates the true identity of this question from a false one.

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