Question

\(\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\) is (a) \(3+\sqrt{5}\) (b) \(3-\sqrt{5}\) (c) \(\frac{3+\sqrt{5}}{2}\) (d) \(\frac{3-\sqrt{5}}{2}\)

Short answer

Question: Compute the value of the following expression: \(\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\) a) \(\frac{3+\sqrt{5}}{2}\) b) \(\frac{3}{2}\) c) \(\frac{1}{\sqrt{3}}\) d) \(\frac{3-\sqrt{5}}{2}\) Answer: (d) \(\frac{3-\sqrt{5}}{2}\)

Step-by-step solution

Analyze the given expression

First, let's take a closer look at the given expression: \(\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\). Our task is to simplify and find its correct value among the multiple choices provided.

Apply the half-angle identity for tangent function

We will use the half-angle identity for tangent function to convert the expression into a more manageable form. The half-angle identity for tangent function is given by: \(\tan\frac{x}{2} = \frac{1-\cos x}{\sin x}\) In our case, \(x = \cos ^{-1} \frac{\sqrt{5}}{3}\). So, the given expression becomes: \(\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right) = \frac{1 - \cos{\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)}}{\sin{\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)}}\)

Simplify the expression

Now, we know that \(\cos\left(\cos^{-1} x\right) = x\). Therefore, \(\cos{\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)} = \frac{\sqrt{5}}{3}\) To find the value of the sine function, we can use the Pythagorean identity, which states: \(\sin^2 x + \cos^2 x = 1\) Now, solving for \(\sin x\): \(\sin{\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)} = \sqrt{1 - \cos^2{\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)}} = \sqrt{1 - \frac{5}{9}} = \frac{2}{3}\)

Substitute the values and find the answer

Substituting the values of sine and cosine in the expression from Step 2: \(\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right) = \frac{1-\frac{\sqrt{5}}{3}}{\frac{2}{3}} =\frac{3-\sqrt{5}}{2}\) Comparing with the given choices, we find that the correct answer is: (d) \(\frac{3-\sqrt{5}}{2}\)

Key concepts

Understanding Half-angle Identities

Trigonometry, the branch of mathematics that deals with the relationships between the sides and angles of triangles, introduces us to various identities that simplify complex problems. One such set of identities are the half-angle identities. These are used to express the trigonometric functions of half angles in terms of the function of the original angle.

For instance, the half-angle identity for the tangent function is represented as
\[\tan\left(\frac{x}{2}\right) = \frac{1 - \cos(x)}{\sin(x)}\]
This identity is particularly useful when we deal with angle measurements that are not readily available in the standard trigonometric tables or when performing integrations in calculus.

In our textbook solution, we cleverly used the half-angle identity to simplify the expression \(\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\). The identity replaced the half-angle with known trigonometric functions, making the expression calculable.

Demystifying Inverse Trigonometric Functions

Inverse trigonometric functions, also known as arcus functions or anti-trigonometric functions, are the inverses of the basic trigonometric functions. They are used to find the angle that corresponds to a given trigonometric ratio. The common inverse trigonometric functions include \(\sin^{-1}\) (arcsin), \(\cos^{-1}\) (arccos), and \(\tan^{-1}\) (arctan).

For example, if you have \(\cos(\theta) = \frac{\sqrt{5}}{3}\), then \(\theta = \cos^{-1}(\frac{\sqrt{5}}{3})\). It means \(\theta\) is the angle whose cosine is \(\frac{\sqrt{5}}{3}\). In our solution, when we encounter \(\cos ^{-1} \frac{\sqrt{5}}{3}\), we're actually looking for an angle whose cosine value is \(\frac{\sqrt{5}}{3}\). By understanding inverse functions, we could further simplify the expression, bringing us one step closer to our solution.

The Essence of the Pythagorean Identity

The Pythagorean identity is one of the fundamental identities in trigonometry, arising directly from the Pythagorean theorem. The most familiar form of this identity is \(\sin^2(x) + \cos^2(x) = 1\), which succinctly expresses the relationship between the sine and cosine of a given angle. This identity holds true for all real values of \(x\).

By utilizing the Pythagorean identity, we can derive one trigonometric function value from another. For example, if the cosine of an angle is known, the sine of the angle can be found, and vice versa. This is exactly what was done in our solution exercise; knowing the cosine enabled us to find the sine necessary to calculate the tangent function. Without the Pythagorean identity, the solution would have been significantly more challenging to obtain. It's a valuable tool for converting between trigonometric functions.