Problem 43 - Algebra and Trigonometry

Question

Write the given expression in terms of x and y only. $$ \cos \left(\sin ^{-1} x-\tan ^{-1} y\right) $$

Step-by-Step Solution

Answer

\( \frac{\sqrt{1-x^2} + xy}{\sqrt{y^2+1}} \)

Step 1: Understand Inverse Trig Functions

The expression \( \cos ( \sin^{-1} x - \tan^{-1} y ) \) involves two inverse trigonometric functions, \( \sin^{-1} x \) and \( \tan^{-1} y \). We aim to simplify this into a form involving only \( x \) and \( y \).

Step 2: Apply the Cosine of a Difference Identity

Utilize the identity for cosine of a difference: \( \cos(a - b) = \cos a \cos b + \sin a \sin b \). Here, let \( a = \sin^{-1} x \) and \( b = \tan^{-1} y \). Substitute these into the identity.

Step 3: Express \( \cos(\sin^{-1} x) \) and \( \sin(\sin^{-1} x) \)

For \( a = \sin^{-1} x \), \( \sin a = x \). Using the identity \( \cos^2 a + \sin^2 a = 1 \), we find \( \cos a = \sqrt{1 - x^2} \).

Step 4: Express \( \cos(\tan^{-1} y) \) and \( \sin(\tan^{-1} y) \)

For \( b = \tan^{-1} y \), consider a right triangle where \( \tan b = \frac{y}{1} \). Therefore, the hypotenuse is \( \sqrt{y^2 + 1} \). We then find \( \sin b = \frac{y}{\sqrt{y^2 + 1}} \) and \( \cos b = \frac{1}{\sqrt{y^2 + 1}} \).

Step 5: Substitute and Simplify

Substituting into the cosine of a difference identity, we have: \( \cos a \cos b + \sin a \sin b = \sqrt{1 - x^2} \frac{1}{\sqrt{y^2 + 1}} + x \frac{y}{\sqrt{y^2 + 1}} \). Simplifying, we get \( \frac{\sqrt{1 - x^2} + xy}{\sqrt{y^2 + 1}} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are a special set of functions that allow us to find angles when given certain trigonometric values. For instance, if you know the sine of an angle equals a particular value, the inverse sine function, written as \( \sin^{-1}(x) \), will help you determine the angle. Similarly, \( \tan^{-1}(y) \) provides the angle whose tangent is \( y \).
These functions are particularly useful in situations where we start with a ratio and we want to find the angle itself, such as when solving trigonometric equations or simplifying complex expressions. It is important to know the range of these inverse functions,

• \( \sin^{-1}(x) \) takes a value between \(-1\) and \(1\) to return an angle from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
• \( \tan^{-1}(y) \), on the other hand, can take any real number and returns an angle between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

Understanding these basics helps us deal effectively with inverse trigonometric values and work through complex trigonometric identities.

Cosine of a Difference Identity

The cosine of a difference identity is a useful tool in trigonometric simplification problems. This identity states that the cosine of a difference can be expressed using the formula: \( \cos(a - b) = \cos a \cos b + \sin a \sin b \).
Using this identity can simplify an expression involving inverse trigonometric functions.
Consider \( a = \sin^{-1} x \) and \( b = \tan^{-1} y \) in our problem. We substitute these values into the identity, transforming the original expression \( \cos(\sin^{-1} x - \tan^{-1} y) \) into a form more manageable with only \( x \) and \( y \).

• The challenge is then to express \( \cos a, \cos b, \sin a, \) and \( \sin b \) in terms of \( x \) and \( y \).

With practice, this identity becomes an invaluable tool for simplifying expressions in trigonometry.

Simplifying Expressions

Simplifying expressions in trigonometry often means breaking down complex expressions into simpler parts that involve only fundamental trigonometric ratios. In our example, we have to use the steps of determining \( \cos a \), \( \cos b \), \( \sin a \), and \( \sin b \) from their inverse function forms.
For \( \sin^{-1} x \), the identity \( \sin a = x \) means that \( \cos a \) can be found using the Pythagorean identity: \( \cos a = \sqrt{1 - x^2} \). This is derived from the fundamental identity \( \cos^2 a + \sin^2 a = 1 \).
For \( \tan^{-1} y \), expressing tangent as opposite over adjacent in a right triangle helps us visualize and find \( \sin b \) and \( \cos b \). With \( \tan b = \frac{y}{1} \), the hypotenuse is \( \sqrt{y^2 + 1} \), thus, \( \sin b = \frac{y}{\sqrt{y^2 + 1}} \) and \( \cos b = \frac{1}{\sqrt{y^2 + 1}} \).

• Combining these findings allows us to simplify the expression using the cosine of a difference identity.

Right Triangle Concept

The right triangle concept is a crucial element in simplifying trigonometric expressions. It allows us to relate trigonometric functions back to a familiar geometric framework. This concept is especially helpful when dealing with inverse trigonometric functions.
Conveniently, any inverse trigonometric function can be conceptualized through a right triangle.
For instance, \( \tan^{-1} y \) can be interpreted as a right triangle where the opposite side is \( y \) and the adjacent side is \( 1 \). This leads directly to the Pythagorean theorem for calculating the hypotenuse and other trigonometric ratios.

• Understanding this visualization helps in easily calculating \( \sin b \) and \( \cos b \) in terms of the sides of the triangle; ultimately simplifying our expression.
• This geometric interpretation makes handling trigonometric identities more intuitive and accessible even for complex problems.

Once mastered, using the right triangle concept becomes a natural approach to solving problems involving trigonometric functions.