Problem 41 - Algebra 2

Question

Find the exact values of the six trigonometric functions of \(\theta\) if the terminal side of \(\theta\) in standard position contains the given point. \((\sqrt{10}, \sqrt{6})\)

Step-by-Step Solution

Answer

The six trigonometric functions are: \(\sin \theta = \frac{\sqrt{6}}{4}\), \(\cos \theta = \frac{\sqrt{10}}{4}\), \(\tan \theta = \frac{\sqrt{15}}{5}\), \(\csc \theta = \frac{2\sqrt{6}}{3}\), \(\sec \theta = \frac{2\sqrt{10}}{5}\), \(\cot \theta = \frac{\sqrt{15}}{3}\).

Step 1: Identify the Coordinates

The point given is \((\sqrt{10}, \sqrt{6})\), where \(x = \sqrt{10}\) and \(y = \sqrt{6}\). This point is on the terminal side of angle \(\theta\) in standard position.

Step 2: Calculate the Radius

The radius, or hypotenuse \(r\), can be found using the Pythagorean theorem:\[r = \sqrt{x^2 + y^2} = \sqrt{(\sqrt{10})^2 + (\sqrt{6})^2} = \sqrt{10 + 6} = \sqrt{16} = 4.\]

Step 3: Find the Sine of \(\theta\)

The sine of \(\theta\) is given by \(\sin \theta = \frac{y}{r}\):\[\sin \theta = \frac{\sqrt{6}}{4}.\]

Step 4: Find the Cosine of \(\theta\)

The cosine of \(\theta\) is given by \(\cos \theta = \frac{x}{r}\):\[\cos \theta = \frac{\sqrt{10}}{4}.\]

Step 5: Find the Tangent of \(\theta\)

The tangent of \(\theta\) is given by \(\tan \theta = \frac{y}{x}\):\[\tan \theta = \frac{\sqrt{6}}{\sqrt{10}} = \frac{\sqrt{6}}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \frac{\sqrt{60}}{10} = \frac{\sqrt{6} \cdot \sqrt{10}}{10}.\] Simplifying further, \(\sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15}\), and thus:\[\tan \theta = \frac{2\sqrt{15}}{10} = \frac{\sqrt{15}}{5}.\]

Step 6: Find the Cosecant of \(\theta\)

The cosecant of \(\theta\) is the reciprocal of the sine:\[\csc \theta = \frac{r}{y} = \frac{4}{\sqrt{6}} = \frac{4}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}.\]

Step 7: Find the Secant of \(\theta\)

The secant of \(\theta\) is the reciprocal of the cosine:\[\sec \theta = \frac{r}{x} = \frac{4}{\sqrt{10}} = \frac{4}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \frac{4\sqrt{10}}{10} = \frac{2\sqrt{10}}{5}.\]

Step 8: Find the Cotangent of \(\theta\)

The cotangent of \(\theta\) is the reciprocal of the tangent:\[\cot \theta = \frac{x}{y} = \frac{\sqrt{10}}{\sqrt{6}} = \frac{\sqrt{10}}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{60}}{6} = \frac{2\sqrt{15}}{6} = \frac{\sqrt{15}}{3}.\]

Key Concepts

Sine

The sine function is all about relating the angle of a right triangle to the ratio of the opposite side over the hypotenuse. In our problem, we are given a point on the terminal side of angle \(\theta\), which is \((\sqrt{10}, \sqrt{6})\). Here, \(y = \sqrt{6}\) acts as the opposite side of the triangle, and \(r = 4\), the hypotenuse or the radius, is calculated using the Pythagorean theorem.

• The formula for sine is \( \sin \theta = \frac{y}{r} \).
• Plugging in the values gives: \( \sin \theta = \frac{\sqrt{6}}{4} \).

This connection helps us understand how sin involves comparing the side opposite an angle to the hypotenuse.

Cosine

Cosine is similar to sine, but instead, it focuses on the adjacent side's relationship to the hypotenuse. For \(\theta\), the adjacent or base is given by \(x = \sqrt{10}\). Just like sine, cosine utilizes the length of the hypotenuse, which is \(r = 4\).

• The formula for cosine is \( \cos \theta = \frac{x}{r} \).
• Substituting the known values results in \( \cos \theta = \frac{\sqrt{10}}{4} \).

Cosine helps us measure how the adjacent side stacks up compared to the hypotenuse in a right triangle.

Tangent

Tangent offers a different view by comparing the opposite side to the adjacent side. It’s a handy ratio especially for dealing with steepness and slopes. For \(\theta\), the opposite side is \(y = \sqrt{6}\) and the adjacent side is \(x = \sqrt{10}\).

• The formula for tangent is \( \tan \theta = \frac{y}{x} \).
• Using our values, \( \tan \theta = \frac{\sqrt{6}}{\sqrt{10}} \), which simplifies to \( \frac{\sqrt{15}}{5} \).

Tangent gives us insight into the angle's slope or how sharp or flat it might be.

Cosecant

Cosecant is a reciprocal function, acting as the inverse of sine. It's useful in specific calculations needing the inversion of the sine relationship. For \(\theta\), where \(r\) is the hypotenuse and \(y\) is the opposite side:

• The formula is \( \csc \theta = \frac{r}{y} \).
• This results in \( \csc \theta = \frac{4}{\sqrt{6}} \), simplifying to \( \frac{2\sqrt{6}}{3} \).

This showcases how we can invert the sine ratio to get cosecant.

Secant

Secant, similar to cosecant, is the reciprocal, but for the cosine function. It gives us the inverse ratio of the cosine's formula, useful in various scenarios requiring this perspective. Here, \(r\) is our hypotenuse and \(x\) is the adjacent side.

• Its formula, \( \sec \theta = \frac{r}{x} \), allows us to calculate \( \sec \theta = \frac{4}{\sqrt{10}} \), simplifying to \( \frac{2\sqrt{10}}{5} \).

Secant helps us switch the cosine perspective into a reciprocal view.

Cotangent

Cotangent flips the classic tangent ratio upside-down, taking the adjacent side to opposite side comparison. In our context, with \(x\) as the adjacent and \(y\) as the opposite, this becomes useful in defining the angle from a new perspective.

• Its formula, \( \cot \theta = \frac{x}{y} \), calculates to give \( \cot \theta = \frac{\sqrt{10}}{\sqrt{6}} \), which simplifies to \( \frac{\sqrt{15}}{3} \).

By understanding cotangent, we see the inverse relationship tangent can provide in a triangle.