Evaluating Inverse Trigonometric Functions Expression

We are asked to find the value of the expression: \(\sin ^{-1} \dfrac{3}{5} - \cos ^{-1} \dfrac{12}{13}\).

This is a common type of mathematics problem involving Inverse Trigonometric Functions. To evaluate this expression, we can use standard trigonometric formulas and properties. One effective way is to convert both terms into the same inverse trigonometric function, for instance, into \(\sin^{-1}\).

Converting \(\cos^{-1} \dfrac{12}{13}\) to \(\sin^{-1}\)

Let \(\theta = \cos ^{-1} \dfrac{12}{13}\). This means \(\cos \theta = \dfrac{12}{13}\). Since the principal value range of \(\cos^{-1} x\) is \([0, \pi]\) and \(\dfrac{12}{13}\) is positive, \(\theta\) lies in \([0, \dfrac{\pi}{2}]\), where \(\sin \theta\) is positive.

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we can find \(\sin \theta\):

\(\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\dfrac{12}{13}\right)^2}\)

\(\sin \theta = \sqrt{1 - \dfrac{144}{169}} = \sqrt{\dfrac{169 - 144}{169}} = \sqrt{\dfrac{25}{169}}\)

\(\sin \theta = \dfrac{5}{13}\) (Taking the positive root as \(\theta \in [0, \dfrac{\pi}{2}]\))

Therefore, \(\theta = \sin ^{-1} \dfrac{5}{13}\). So, \(\cos ^{-1} \dfrac{12}{13} = \sin ^{-1} \dfrac{5}{13}\).

Using the \(\sin^{-1} x - \sin^{-1} y\) Formula

Now the original expression becomes \(\sin ^{-1} \dfrac{3}{5} - \sin ^{-1} \dfrac{5}{13}\).

We can use the trigonometric formulas for the difference of two sin inverse functions. The formula is:

\(\sin^{-1} x - \sin^{-1} y = \sin^{-1} \left(x \sqrt{1 - y^2} - y \sqrt{1 - x^2}\right)\), where \(|x| \le 1\), \(|y| \le 1\).

In our case, \(x = \dfrac{3}{5}\) and \(y = \dfrac{5}{13}\). Both are between 0 and 1, so the formula applies.

First, calculate the terms under the square roots:

• \(\sqrt{1 - x^2} = \sqrt{1 - \left(\dfrac{3}{5}\right)^2} = \sqrt{1 - \dfrac{9}{25}} = \sqrt{\dfrac{25 - 9}{25}} = \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}\)
• \(\sqrt{1 - y^2} = \sqrt{1 - \left(\dfrac{5}{13}\right)^2} = \sqrt{1 - \dfrac{25}{169}} = \sqrt{\dfrac{169 - 25}{169}} = \sqrt{\dfrac{144}{169}} = \dfrac{12}{13}\)

Now substitute these values into the formula:

\(\sin^{-1} \left(\dfrac{3}{5} \cdot \dfrac{12}{13} - \dfrac{5}{13} \cdot \dfrac{4}{5}\right)\)

Perform the multiplication:

\(\sin^{-1} \left(\dfrac{3 \times 12}{5 \times 13} - \dfrac{5 \times 4}{13 \times 5}\right) = \sin^{-1} \left(\dfrac{36}{65} - \dfrac{20}{65}\right)\)

Finally, perform the subtraction:

\(\sin^{-1} \left(\dfrac{36 - 20}{65}\right) = \sin^{-1} \dfrac{16}{65}\)

Conclusion

The value of the expression \(\sin ^{-1} \dfrac{3}{5} - \cos ^{-1} \dfrac{12}{13}\) is \(\sin^{-1} \dfrac{16}{65}\). This demonstrates how Inverse Trigonometric Functions problems can be solved using appropriate trigonometric identities and formulas. Mastering these Inverse Trigonometric Functions concepts is crucial for solving such mathematics problems.