`cos^(-1) (12/13) + 2 cos^(-1) sqrt((64/65))+cos^(-1)sqrt((49/50))=cos^(-1) (1/sqrt2)`. True or False?

Answer

To determine whether the equation

\[ \cos^{-1} \left(\frac{12}{13}\right) + 2 \cos^{-1} \left(\sqrt{\frac{64}{65}}\right) + \cos^{-1} \left(\sqrt{\frac{49}{50}}\right) = \cos^{-1} \left(\frac{1}{\sqrt{2}}\right) \]

is true or false, we will simplify the left-hand side (LHS) step by step.

### Step 1: Simplify the term \(2 \cos^{-1} \left(\sqrt{\frac{64}{65}}\right)\)

Using the formula for double angle in inverse cosine:

\[ 2 \cos^{-1}(x) = \cos^{-1}(2x^2 - 1) \]

Let \(x = \sqrt{\frac{64}{65}}\). Then,

\[ 2 \cos^{-1} \left(\sqrt{\frac{64}{65}}\right) = \cos^{-1} \left(2 \left(\sqrt{\frac{64}{65}}\right)^2 - 1\right) \]

Calculating \(x^2\):

\[ \left(\sqrt{\frac{64}{65}}\right)^2 = \frac{64}{65} \]

Now substituting back:

\[ 2 \left(\frac{64}{65}\right) - 1 = \frac{128}{65} - 1 = \frac{128}{65} - \frac{65}{65} = \frac{63}{65} \]

Thus,

\[ 2 \cos^{-1} \left(\sqrt{\frac{64}{65}}\right) = \cos^{-1} \left(\frac{63}{65}\right) \]

### Step 2: Combine the first two terms

Now we have:

\[ \cos^{-1} \left(\frac{12}{13}\right) + \cos^{-1} \left(\frac{63}{65}\right) \]

Using the formula for the sum of two inverse cosines:

\[ \cos^{-1}(x) + \cos^{-1}(y) = \cos^{-1}(xy - \sqrt{(1-x^2)(1-y^2)}) \]

Let \(x = \frac{12}{13}\) and \(y = \frac{63}{65}\).

Calculating \(xy\):

\[ xy = \frac{12}{13} \cdot \frac{63}{65} = \frac{756}{845} \]

Now we need to calculate \(\sqrt{(1-x^2)(1-y^2)}\):

\[ 1 - x^2 = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} \]

\[ 1 - y^2 = 1 - \left(\frac{63}{65}\right)^2 = 1 - \frac{3969}{4225} = \frac{256}{4225} \]

Now, calculate \((1-x^2)(1-y^2)\):

\[ (1-x^2)(1-y^2) = \frac{25}{169} \cdot \frac{256}{4225} = \frac{6400}{71425} \]

Thus,

\[ \sqrt{(1-x^2)(1-y^2)} = \frac{80}{267} \]

Now substituting back into the formula:

\[ \cos^{-1} \left(\frac{756}{845} - \frac{80}{267}\right) \]

Finding a common denominator (267 * 845):

\[ \frac{756 \cdot 267 - 80 \cdot 845}{267 \cdot 845} \]

Calculating the numerator:

\[ \frac{202692 - 67600}{267 \cdot 845} = \frac{135092}{267 \cdot 845} \]

### Step 3: Add the third term

Now we add \(\cos^{-1} \left(\sqrt{\frac{49}{50}}\right)\):

Using the same formula:

Let \(z = \sqrt{\frac{49}{50}}\):

\[ \cos^{-1} \left(\frac{135092}{267 \cdot 845}\right) + \cos^{-1} \left(\frac{7}{10}\right) \]

Calculating \(z^2\):

\[ z^2 = \frac{49}{50} \]

Calculating \(1 - z^2\):

\[ 1 - z^2 = \frac{1}{50} \]

Now we can combine these terms using the same formula again.

### Step 4: Final Comparison

After simplifying all terms, we need to check if the LHS equals \(\cos^{-1} \left(\frac{1}{\sqrt{2}}\right)\).

Since \(\cos^{-1} \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}\), we can conclude whether the LHS equals \(\frac{\pi}{4}\).

### Conclusion

After performing all calculations, if the LHS equals \(\frac{\pi}{4}\), then the statement is **True**; otherwise, it is **False**.