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Chapter 8: Problem 34

Find the exact value of each expression. $$ \sin \left(\cos ^{-1} \frac{1}{2}\right) $$

Short Answer

Expert verified
\(\sin \left( \cos^{-1} \frac{1}{2} \right) = \frac{\sqrt{3}}{2}\)

Step by step solution

01

Understand the inverse function

The expression \(\cos^{-1} \frac{1}{2}\) represents the angle \(\theta\) such that \(\cos \theta = \frac{1}{2}\). First, identify this angle.
02

Identify the angle

Recall that \(\cos \theta = \frac{1}{2}\) for \(\theta = \frac{\pi}{3}\). Hence, \(\cos^{-1} \frac{1}{2} = \frac{\pi}{3}\).
03

Find the sine of the angle

To find \(\sin \left(\cos^{-1} \frac{1}{2}\right)\), substitute \(\cos^{-1} \frac{1}{2}\) with \(\frac{\pi}{3}\). Thus, \(\sin \left(\cos^{-1} \frac{1}{2}\right) = \sin \frac{\pi}{3}\).
04

Determine the exact sine value

Recall the exact value of \(\sin \frac{\pi}{3}\). \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Trigonometric Functions
Inverse trigonometric functions help us find the angle when we know the value of a trigonometric function. For example, \(\cos^{-1}\left(\frac{1}{2}\right)\) asks for the angle \(\theta\) such that \(\theta\) satisfies \(\cos \theta = \frac{1}{2}\). The result is an angle because the inverse function reverses the original trigonometric function.
These functions have specific ranges:
  • \(\cos^{-1}(x)\) gives an angle in the range [0, \(\pi\)]
  • \(\sin^{-1}(x)\) gives an angle in the range [鈭抃(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
  • \(\tan^{-1}(x)\) gives an angle in the range [鈭抃(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]

Understanding inverse trigonometric functions is crucial for solving problems where you need to identify an angle based on known function values.
Cosine Function
The cosine function, often abbreviated as \(\cos\), is a primary trigonometric function. It relates the adjacent side to the hypotenuse in a right triangle. Mathematically, for any angle \(\theta\), cosine is defined as:
\[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \]
  • Cosine values range from 鈭�1 to 1.
  • The cosine of 0 degrees (or 0 radians) is 1.
  • The function is periodic with a period of \(2\pi\), meaning \(\cos(\theta + 2\pi) = \cos \theta\).

In the unit circle, the cosine of an angle is the x-coordinate of the point where the terminal side of the angle intersects the circle. For example, \(\cos \frac{\pi}{3} = \frac{1}{2}\), indicating that the x-coordinate at this angle is \(\frac{1}{2}\).
Sine Function
The sine function, abbreviated as \(\sin\), is another fundamental trigonometric function. It relates the opposite side to the hypotenuse in a right triangle. Formally, for an angle \(\theta\):
\[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \]
  • Sine values also range from 鈭�1 to 1.
  • \(\sin(0) = 0\) and \(\sin(\frac{\pi}{2}) = 1\).
  • Like cosine, the sine function is periodic with a period of \(2\pi\).

In the unit circle, the sine of an angle corresponds to the y-coordinate of the intersecting point on the circle. To find the sine of \(\cos^{-1}(\frac{1}{2})\), recognize that if \(\cos \frac{\pi}{3} = \frac{1}{2}\), then \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\).
Unit Circle
The unit circle is a circle with a radius of 1 centered at the origin of a coordinate system. It's a powerful tool in trigonometry for understanding angles and their corresponding sine and cosine values. The equation of the unit circle is:
\[ x^2 + y^2 = 1 \]
Here are some notable points on the unit circle:
  • At \(0\) radians, \(\cos(0) = 1\) and \(\sin(0) = 0\).
  • At \(\frac{\pi}{2}\) radians, \(\cos(\frac{\pi}{2}) = 0\) and \(\sin(\frac{\pi}{2}) = 1\).
  • At \(\frac{\pi}{3}\) radians, \(\cos(\frac{\pi}{3}) = \frac{1}{2}\) and \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\).
  • The unit circle helps in visualizing the relationships and values of trigonometric functions for various angles.

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Most popular questions from this chapter

Write each trigonometric expression as an algebraic expression containing u and \(v .\) Give the restrictions required on \(u\) and \(v\). $$ \tan \left(\sin ^{-1} u-\cos ^{-1} v\right) $$

Graph \(f(x)=\sin ^{2} x=\frac{1-\cos (2 x)}{2}\) for \(0 \leq x \leq 2 \pi\) by using transformations.

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If the angle of incidence and the angle of refraction are complementary angles, the angle of incidence is referred to as the Brewster angle \(\theta_{B}\). The Brewster angle is related to the indices of refraction of the two media, \(n_{1}\) and \(n_{2},\) by the equation \(n_{1} \sin \theta_{B}=n_{2} \cos \theta_{B},\) where \(n_{1}\) is the index of refraction of the incident medium and \(n_{2}\) is the index of refraction of the refractive medium. Determine the Brewster angle for a light beam traveling through water (at \(20^{\circ} \mathrm{C}\) ) that makes an angle of incidence with a smooth, flat slab of crown glass.

Use the following discussion. The formula $$ D=24\left[1-\frac{\cos ^{-1}(\tan i \tan \theta)}{\pi}\right] $$ Approximate the number of hours of daylight for any location that is \(66^{\circ} 30^{\prime}\) north latitude for the following dates: (a) Summer solstice \(\left(i=23.5^{\circ}\right)\) (b) Vernal equinox \(\left(i=0^{\circ}\right)\) (c) July \(4\left(i=22^{\circ} 48^{\prime}\right)\) (d) Thanks to the symmetry of the orbital path of Earth around the Sun, the number of hours of daylight on the winter solstice may be found by computing the number of hours of daylight on the summer solstice and subtracting this result from 24 hours. Compute the number of hours of daylight for this location on the winter solstice. What do you conclude about daylight for a location at \(66^{\circ} 30^{\prime}\) north latitude?
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