Question

Right-triangle relationships Draw a right triangle to simplify the given expressions. Assume \(x>0.\) $$\sin \left(\cos ^{-1}(x / 2)\right)$$

Short answer

Answer: The simplified expression is \(\frac{\sqrt{4 - x^2}}{2}\).

Step-by-step solution

Draw a right triangle

Draw a right triangle with one angle, labeled as \(\theta\). We'll use \(\cos ^{-1}(x / 2)\) to represent this angle. The side adjacent to \(\theta\) can be denoted as length \(x\), and the hypotenuse is of length 2 since we have \(\cos \theta = \frac{x}{2}\).

Find the length of the remaining side

We can now determine the length of the remaining side of the triangle by using the Pythagorean theorem. Let's denote the length of this side as \(y\). We have the equation: $$y^2 + x^2 = 2^2$$ Solve for \(y\): $$y = \sqrt{4 - x^2}$$

Apply the sine function to the triangle

Now that we know the length of all sides of the triangle, we can use the sine function, which is given by the ratio of the opposite side to the hypotenuse. In this case, \(y\) represents the opposite side and 2 is the hypotenuse. We have: $$\sin \theta = \frac{y}{2}$$ Substitute the value of \(y\) from step 2: $$\sin \theta = \frac{\sqrt{4-x^2}}{2}$$

Substitute the value of \(\theta\)

Since \(\theta = \cos ^{-1}(x / 2)\), substitute this into the equation from step 3: $$\sin \left(\cos ^{-1}(x / 2)\right) = \frac{\sqrt{4 - x^2}}{2}$$ The final simplified expression is: \(\frac{\sqrt{4 - x^2}}{2}\).

Key concepts

Right Triangle

A right triangle is a triangle in which one of its angles is exactly 90 degrees. This angle makes it unique and allows for various trigonometric relationships to emerge. In a right triangle, the side opposite the right angle is the longest and is called the hypotenuse. The other two sides are known as the legs of the triangle.

When dealing with right triangles, we often label one of the angles as \(\theta\). From this specific angle, the side directly opposite is called the "opposite" side, and the side adjacent to it, but not the hypotenuse, is the "adjacent" side. These relationships are very important when it comes to understanding trigonometric functions, as they define how the sides relate to one another through these functions.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to determine the angle when you know the sides of a right triangle. They are essentially the opposite of the regular trigonometric functions. For example, while the cosine function tells us the ratio of the adjacent side to the hypotenuse of an angle in a right triangle, \(\cos^{-1}\) (or arccos) allows us to find that angle when given the ratio.

• \(\sin^{-1}(x)\): Finds the angle whose sine is \(x\).
• \(\cos^{-1}(x)\): Finds the angle whose cosine is \(x\).
• \(\tan^{-1}(x)\): Finds the angle whose tangent is \(x\).

In our example, \(\cos ^{-1}(x / 2)\), helps us identify the angle of the right triangle for which the adjacent side to hypotenuse ratio is \(\frac{x}{2}\). This setup allows us to further explore other relationships within the triangle.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in trigonometry and geometry that applies solely to right triangles. It describes the relationship between the lengths of the sides of a right triangle. The theorem states that "the square of the hypotenuse is equal to the sum of the squares of the other two sides." Mathematically, this is expressed as:
\[a^2 + b^2 = c^2\]
Here, \(c\) represents the hypotenuse, and \(a\) and \(b\) are the triangle's other two sides.

In practical terms, if you know the lengths of any two sides of a right triangle, the Pythagorean Theorem lets you calculate the length of the third side. In our problem, it allows us to find the opposite side \(y\) when we know \(x\) and the hypotenuse. By setting up the equation:
\[y^2 + x^2 = 4\]
We can solve for \(y\) to get:
\[y = \sqrt{4 - x^2}\]
This is crucial in our calculation of the sine function in the given problem. Understanding and using the Pythagorean Theorem is key to solving many problems involving right triangles.