Problem 32 - Algebra 2 and Trigonometry

Question

In \(24-32,\) find the exact value of each expression. $$ \cos \left(\arcsin \left(-\frac{\sqrt{2}}{2}\right)\right) $$

Step-by-Step Solution

Answer

The exact value is \( \frac{\sqrt{2}}{2} \).

Step 1: Understand Arcsin Result

Recall that the function \( \arcsin(x) \) returns an angle \( \theta \) such that \( \sin(\theta) = x \). Therefore, we have \( \sin(\theta) = -\frac{\sqrt{2}}{2} \). Since \( \arcsin(x) \) gives angles in the range \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\), the angle \( \theta \) must be \( -\frac{\pi}{4} \) because \( \sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \).

Step 2: Find Cosine of the Angle

Now we need to find \( \cos(\theta) \) where \( \theta = -\frac{\pi}{4} \). Using the cosine value for this standard angle, we know \( \cos(-\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \), since cosine is an even function \( \cos(-x) = \cos(x) \).

Step 3: Conclude the Expression

We have determined that \( \theta = -\frac{\pi}{4} \) leads to \( \cos(\theta) = \frac{\sqrt{2}}{2} \). Therefore, the exact value of the expression \( \cos \left( \arcsin \left( -\frac{\sqrt{2}}{2} \right) \right) \) is \( \frac{\sqrt{2}}{2} \).

Key Concepts

Understanding the Arcsin Function

The arcsin function, known as the inverse sine function, is used to find the angle when the value of the sine is given. If you have a value like \(-\frac{\sqrt{2}}{2}\), simply apply the arcsin function to find out which angle has that sine value. This function is useful for determining specific angles in trigonometry, especially when you're dealing with right triangles or unit circles.

When you apply \(\arcsin(x)\), the output is an angle \(\theta\) such that \(\sin(\theta) = x\). For values between -1 and 1, the function provides precise angles within the range \(-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\). This means that any output from the arcsin function will represent an angle in this interval.

• Example: \(\arcsin\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}\)
• Reason: The sine of \(-\frac{\pi}{4}\) is \(-\frac{\sqrt{2}}{2}\), fitting within the defined range of the arcsin function.

Finding the Cosine of an Angle

Once you've found an angle using the arcsin function, calculating the cosine of that angle involves understanding trigonometric relationships on the unit circle or within triangles. The cosine function corresponds to the horizontal position of a point on the unit circle.

In our example, after finding \(\theta = -\frac{\pi}{4}\) from \(\arcsin(\sin(\theta))\), we're tasked with finding \(\cos(\theta)\). From trigonometric tables or the unit circle, we know that:

• \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\)
• This identity holds true even when dealing with negative angles, due to cosine's properties.

Thus, \(\cos\left(-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\) because cosine is an even function. This simplifies finding cosine for angles like \(-\frac{\pi}{4}\).

Exploring the Even Function Property

An even function has a unique characteristic where the function's value remains unchanged if \(x\) is replaced by \(-x\). In the world of trigonometry, cosine exemplifies this property perfectly. This means for any angle \(\theta\):

• \(\cos(-\theta) = \cos(\theta)\)
• This property is especially beneficial when calculating trigonometric functions of negative angles.

What makes this intriguing is how it explains why the cosine of \(-\frac{\pi}{4}\) equals the cosine of \(\frac{\pi}{4}\), both computed as \(\frac{\sqrt{2}}{2}\). This property greatly simplifies problems involving negative angles.

In summary, recognizing that cosine is an even function can save time and reduce errors when solving trigonometric equations, particularly when negative angles are involved.