Q.Find the value of the following: $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$

Concept: Inverse Cosine Addition — use the formula $\cos^{-1}x + \cos^{-1}y = \cos^{-1}\left(xy - \sqrt{1-x^2}\sqrt{1-y^2}\right)$ when the sum lies in $[0, \pi]$.

Step 1: Let $\alpha = \cos^{-1}\frac{4}{5}$ and $\beta = \cos^{-1}\frac{12}{13}$. Then $\cos\alpha = \frac{4}{5}$, $\cos\beta = \frac{12}{13}$, and both angles are in $[0, \pi/2]$ since the cosines are positive.

Step 2: Compute $\sin\alpha = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$, $\sin\beta = \sqrt{1 - \frac{144}{169}} = \frac{5}{13}$.

Step 3: Using the addition formula: …

We need to verify that $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$.

The core idea: when you have a sum of two inverse trigonometric functions, you can't just add the arguments. Instead, take the cosine of both sides. If we can show that the cosine of the left-hand side equals $\frac{33}{65}$, and that the left-hand side lies in the range where $\cos^{-1}$ is one-to-one (i.e., $[0,\pi]$), then the equality is proven.

Let $A = \cos^{-1} \frac{4}{5}$ and $B = \cos^{-1} \frac{12}{13}$. Then $\cos A = \frac{4}{5}$ and $\cos B = \frac{12}{13}$. Since $\cos^{-1}$ returns angles in $[0,\pi]$, both $A$ and $B$ are in that interval. Their sum $A+B$ could be up to $2\pi$, but we'll check its actual range.

We need $\cos(A+B)$. The cosine addition formula is:

We know $\cos A$ and $\cos B$, but need $\sin A$ and $\sin B$. Since $A,B \in [0,\pi]$, sine is non-negative in this interval (sine is positive for angles in $(0,\pi)$, zero only at endpoints). So we can take the positive square root.

1. Find $\sin A$

   $\sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}$

   Since $A \in [0,\pi]$, $\sin A \ge 0$, so $\sin A = \frac{3}{5}$.

2. Find $\sin B$

   $\sin^2 B = 1 - \cos^2 B = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}$

   So $\sin B = \frac{5}{13}$ (positive for the same reason).

3. Apply the cosine addition formula

So $\cos(A+B) = \frac{33}{65}$. This means $A+B$ is an angle whose cosine is $\frac{33}{65}$. But $\cos^{-1}\frac{33}{65}$ is the angle in $[0,\pi]$ with that cosine. So if $A+B$ itself lies in $[0,\pi]$, then $A+B = \cos^{-1}\frac{33}{65}$.

4. Check the range of $A+B$ …

Method: Combining a sum of inverse cosines

Use this to verify or evaluate $\cos^{-1}x + \cos^{-1}y$ as a single inverse cosine.

Steps

Step 1: Name each term as an angle.

Set $A = \cos^{-1}x$ and $B = \cos^{-1}y$, so $\cos A = x$, $\cos B = y$, with both angles in $[0,\pi]$.

Step 2: Take the cosine of the sum.

Sines are non-negative here because $A, B \in [0,\pi]$, so use positive roots. …

Common Mistakes

Mistake 1: Taking $\sin A$ or $\sin B$ negative.

Why it's wrong: $A = \cos^{-1}\frac{4}{5}$ and $B = \cos^{-1}\frac{12}{13}$ lie in $[0,\pi]$, where sine is non-negative, so $\sin A = +\frac{3}{5}$ and $\sin B = +\frac{5}{13}$. Correct approach: always use the positive root for $\sqrt{1-x^2}$ here.

Mistake 2: Applying $\cos^{-1}x + \cos^{-1}y = \cos^{-1}(\cdots)$ without checking $x+y$. …