Question

Draw a right triangle to simplify the given expressions. $$\cos \left(\sec ^{-1} x\right)$$

Short answer

Answer: The expression for the cosine of the inverse secant function is $$\frac{1}{x}$$.

Step-by-step solution

Defining the inverse secant function

The inverse secant function, \(\sec^{-1}x\), is an angle whose secant is x, which means that: $$\sec \left(\sec ^{-1} x\right) = x$$ Since secant is the reciprocal of the cosine function, this can be written as: $$\frac{1}{\cos \left(\sec ^{-1} x\right)} = x$$

Drawing the right triangle

Now, let's draw a right triangle with one of its acute angles representing \(\sec^{-1}x\). Let's call that angle A. Since \(\sec \left(\sec ^{-1} x\right) = x\), in this right triangle, the ratio of the hypotenuse to the adjacent side will be x. Let the adjacent side be of length a, and the hypotenuse be of length x*a, as the ratio of the hypotenuse and adjacent side is x. This triangle now has an angle A, with adjacent side a and hypotenuse x*a.

Calculating the length of the opposite side using Pythagorean Theorem

To find the length of the remaining side (opposite side) in the triangle, we will use the Pythagorean theorem: $$(\text{hypotenuse})^2 = (\text{adjacent side})^2 + (\text{opposite side})^2$$ Substituting the known values, we have: $$(x*a)^2 = a^2 + (\text{opposite side})^2$$ Let's call the opposite side b. Then the equation becomes: $${\left( x*a \right)}^2 = a^2 + b^2$$ Now, we rearrange the equation and solve for b: $$b^2 = (x*a)^2 - a^2$$ $$b=\sqrt{(x*a)^2-a^2}$$

Finding the expression for cosine of the inverse secant function

In the context of the right triangle, the cosine function can be defined as the ratio of the adjacent side to the hypotenuse: $$\cos A = \frac{\text{adjacent side}}{\text{hypotenuse}}$$ So, $$\cos \left(\sec ^{-1} x\right) = \frac{a}{x*a}$$ From Step 3, we have the expression for opposite side (b), which will be used to find the value of a. In the triangle, we know that: $$\sin A = \frac{\text{opposite side}}{\text{hypotenuse}}$$ So, $$\sin \left(\sec ^{-1} x\right) = \frac{\sqrt{(x*a)^2-a^2}}{x*a}$$ Now, using the Pythagorean identity $$\sin^2(\theta) + \cos^2(\theta) = 1$$, we can find the value of a. For angle A, we have: $$\sin^2\left(\sec ^{-1} x\right) + {\left( \frac{a}{x*a} \right)}^2 = 1$$ Substituting the expression for sin from Step 3: $$\left(\frac{\sqrt{(x*a)^2-a^2}}{x*a}\right)^2 + {\left( \frac{a}{x*a} \right)}^2 = 1$$ Now, solve for a in the equation: $$a=\frac{1}{\sqrt{x^2-1}}$$

Final expression for cosine of the inverse secant function

Now we can substitute the value of a back into the expression for cosine: $$\cos \left(\sec ^{-1} x\right) = \frac{a}{x*a} = \frac{\frac{1}{\sqrt{x^2-1}}}{x*\frac{1}{\sqrt{x^2-1}}}$$ Simplifying the expression gives: $$\cos \left(\sec ^{-1} x\right) = \frac{1}{x}$$ Therefore, the expression $$\cos \left(\sec ^{-1} x\right)$$ simplifies to $$\frac{1}{x}$$ when considering the right triangle with angle A representing $$\sec ^{-1} x$$.

Key concepts

Right Triangle Trigonometry

Understanding right triangle trigonometry requires knowledge about the relationships between the angles and sides of right triangles. In such triangles, one angle is always 90 degrees. The other two angles are acute and their trigonometric relationships are key to solving many problems.
For any right triangle:

• The hypotenuse is the longest side, opposite the right angle.
• The adjacent side is next to the angle of interest but not the hypotenuse.
• The opposite side is the side across from the angle of interest.

Trigonometric functions such as sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)) relate these triangles’ sides to their angles.

• \(\sin (\text{angle}) = \frac{\text{opposite side}}{\text{hypotenuse}}\)
• \(\cos (\text{angle}) = \frac{\text{adjacent side}}{\text{hypotenuse}}\)
• \(\tan (\text{angle}) = \frac{\text{opposite side}}{\text{adjacent side}}\)

Since the exercise requires finding \(\cos(\sec^{-1} x)\), we use the inverse trigonometric function to determine the angle whose secant is \(x\).Then we construct a right triangle to locate the cosine value for this specific angle.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry that applies only to right triangles. It states that the square of the length of the hypotenuse (\(c\)) is equal to the sum of the squares of the lengths of the other two sides (\(a\) and \(b\)):
\[c^2 = a^2 + b^2\]
This theorem is essential when finding the unknown side length in right triangle problems, especially when the lengths of two sides are known.
In the exercise, to solve for the missing side of the triangle, we rearranged the Pythagorean theorem formula to find \(b\) (opposite side) using:
\[b = \sqrt{(x \, \cdot \, a)^2 - a^2}\]
Such rearrangements allow us to apply known information—here the lengths of the hypotenuse (\(x\cdot a\)) and adjacent side (\(a\))—to find the remaining side’s length. This is important for using trigonometric identities to solve for other trigonometric expressions.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for any value of the involved variables. These identities are extremely useful tools for simplifying expressions and solving trigonometric equations.
Commonly used identities include:

• Pythagorean identities, most notably \(\sin^2(\theta) + \cos^2(\theta) = 1\)
• Reciprocal identities, like \(\sec(\theta) = \frac{1}{\cos(\theta)}\)

In our exercise, we used the Pythagorean identity to solve for the cosine of an angle derived from the inverse secant function. By finding the sine of the angle first and applying the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\), we were able to calculate the cosine value, thereby simplifying the expression to \(\frac{1}{x}\).
Trigonometric identities not only simplify calculations but also unveil the underlying relationships between trigonometric functions, making them invaluable in various mathematical and engineering applications.