Miscellaneous · Q5: $\cos^{-1}\frac45+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$

Basics you need

Solution

1. Name the two angles and locate them

   Let

   $$A = \cos^{-1}\frac45, \qquad B = \cos^{-1}\frac{12}{13}.$$

   Then $\cos A = \frac45$⁠, $\cos B = \frac{12}{13}$⁠, and $A$⁠, $B$ are in $[0,\pi]$⁠. Both cosines are positive, which rules out $\left[\frac{\pi}{2},\pi\right]$⁠; both are less than 1, which rules out $0$⁠. So

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

   We must prove $A + B = \cos^{-1}\frac{33}{65}$⁠.

   Naming the angles lets us use trigonometry on them. Knowing they are acute lets us draw them in right triangles, and gives the bound on $A+B$ needed in Step 4.
   $y = \cos^{-1}x$ means $\cos y = x$ and $y \in [0,\pi]$ (Basic 1).

2. Build the two triangles

   For $A$⁠: adjacent 4, hypotenuse 5, so the opposite side is $\sqrt{25 - 16} = 3$⁠. For $B$⁠: adjacent 12, hypotenuse 13, so the opposite side is

   $$\sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5.$$

   Hence

   $$\sin A = \frac35, \qquad \sin B = \frac{5}{13}.$$
   The addition formula for $\cos(A+B)$ needs the sines as well as the cosines. From $\sin^2 A = 1 - \cos^2 A = \frac{9}{25}$ we get $\sin A = \pm\frac35$⁠, and the positive root is correct because $A \in [0,\pi]$⁠, where sine is never negative (Basic 4). The same holds for $B$⁠.
   $3^2 + 4^2 = 5^2$ and $5^2 + 12^2 = 13^2$⁠. Lesson: the triangle trick

3. Add the angles: $\cos(A+B) = \frac{33}{65}$

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

   For comparison, the sine of the sum is

   $$\begin{aligned} \sin(A+B) &= \sin A\cos B + \cos A\sin B \\[1mm] &= \tfrac{36}{65} + \tfrac{20}{65} = \tfrac{56}{65}, \end{aligned}$$

   so $A+B$ belongs to a 33, 56, 65 triangle (check: $33^2 + 56^2 = 1089 + 3136 = 4225 = 65^2$⁠).

   The right-hand side of the question is a $\cos^{-1}$⁠, so we compute the cosine of the left-hand side. The common denominator $5 \times 13 = 65$ is why the answer has 65 at the bottom.
   $\cos(A+B) = \cos A\cos B - \sin A\sin B$ (Class XI, Chapter 3).

4. The crucial check: $A+B$ lies in $[0,\pi]$

   Step 3 shows that $A+B$ and $\cos^{-1}\frac{33}{65}$ have the same cosine. They are the same angle only if $A+B$ lies in $[0,\pi]$⁠, because $\cos^{-1}\frac{33}{65}$ is the only angle in $[0,\pi]$ with cosine $\frac{33}{65}$⁠.

   Add the two inequalities from Step 1:

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

   So $A+B \in [0,\pi]$⁠. (In fact $\cos(A+B) > 0$ shows it is acute, about $59.5^\circ$⁠, but we do not need that here.)

   Many angles have cosine $\frac{33}{65}$⁠: $\pm(A+B) + 2k\pi$ for every integer $k$ (positive, negative or zero). Only one of them is in $[0,\pi]$⁠, and that one is $\cos^{-1}\frac{33}{65}$⁠. So the inequality $0 < A+B < \pi$ is what turns "same cosine" into "same angle". Here it is easy because both angles are acute; Step 6 shows a case where it fails.
   $\cos^{-1}(\cos x) = x$ for $x \in [0,\pi]$ (Basic 5).

5. Conclude

   $A+B$ lies in $[0,\pi]$ (Step 4) and $\cos(A+B) = \frac{33}{65}$ (Step 3). Therefore

   $$\begin{aligned} \cos^{-1}\frac{33}{65} &= \cos^{-1}\bigl(\cos(A+B)\bigr) = A + B \\[1mm] &= \cos^{-1}\frac45 + \cos^{-1}\frac{12}{13}. \end{aligned}$$

   This is the required result.

   Note that $\cos^{-1}\frac45$ is the same angle as $\sin^{-1}\frac35$ (the 3, 4, 5 triangle again). So the left side here equals the left side of Miscellaneous Q6, and together the two questions say $\cos^{-1}\frac{33}{65} = \sin^{-1}\frac{56}{65}$⁠. That is one and the same angle of the 33, 56, 65 triangle, described once by its cosine and once by its sine.

6. Why the range check is not a formality

   Change both cosines to negative: $A' = \cos^{-1}\left(-\frac45\right)$ and $B' = \cos^{-1}\left(-\frac{12}{13}\right)$⁠. These are obtuse: $A' = \pi - A$ and $B' = \pi - B$⁠. The same formula gives

   $$\begin{aligned} \cos(A'+B') &= \left(-\tfrac45\right)\left(-\tfrac{12}{13}\right) - \tfrac35\cdot\tfrac{5}{13} \\[1mm] &= \tfrac{33}{65}. \end{aligned}$$

   But $A' + B' = 2\pi - (A+B) \approx 5.2449$⁠, which is bigger than $\pi$⁠. So $A' + B' \ne \cos^{-1}\frac{33}{65} \approx 1.0383$⁠; instead $A' + B' = 2\pi - \cos^{-1}\frac{33}{65}$⁠.

   The cosine computation cannot tell $A+B$ from $2\pi - (A+B)$⁠. Only the range check can.

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