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Find exact values for each of the following. (No calculator-or use it only to check your answers.) (a) \(\cos (\pi / 4)\) (b) \(\cos (5 \pi / 4)\) (c) \(\cos (-3 \pi / 4)\) (d) \(\sin (5 \pi / 6)\) (e) \(\sin (-13 \pi / 6)\) (f) \(\cos (-2 \pi / 3)\)

Short Answer

Expert verified
(a) \(\sqrt{2}/2\) (b) -\(\sqrt{2}/2\) (c) -\(\sqrt{2}/2\) (d) 1/2 (e) 1/2 (f) -1/2

Step by step solution

01

Find the values of cosine

Start by finding the values for cosine. We'll use the unit circle to do this. Recall that in the unit circle, \(\cos(\theta)\) gives the x-coordinate of the point on the unit circle. (a) From the unit circle, we can see that \(\cos(\pi / 4)\) is \(\sqrt{2}/2\). (b) For \(\cos(5\pi /4)\), the angle corresponds to the third quadrant where cosine is negative, so \(\cos(5\pi /4)\) = -\(\sqrt{2}/2\). (c) For \(\cos(-3\pi /4)\), negative angle means we go in the clockwise direction. So \ (-3\pi /4)\ corresponds to the second quadrant where cosine is also negative, so \(\cos(-3\pi /4)\) = -\(\sqrt{2}/2\). (f) Similarly, \ (-2\pi /3)\ corresponds to the second quadrant where cosine is also negative, so \(\cos(-2\pi /3)\) = -\(1/2\).
02

Find the values of sine

Now let's find the values for sine. Again, we'll use the unit circle. Recall that in the unit circle, \(\sin(\theta)\) gives the y-coordinate of the point on the unit circle. (d) From the unit circle, we can see that \(\sin(5\pi /6)\) is 1/2. (e) For \(\sin(-13\pi /6)\), the negative angle means we go in the clockwise direction. Notice that \ (-13\pi /6)\ is equivalent to \(\pi /6\) in the unit circle, so \(\sin(-13\pi /6)\) = 1/2.

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Key Concepts

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

Cosine Function
The cosine function is a fundamental component of trigonometry and is essential in understanding the behavior of angles in a unit circle. The cosine of an angle gives us the x-coordinate of a point that lies on the circumference of the unit circle, which is a circle with a radius of 1 unit centered at the origin of a coordinate system.

In practical terms, to find the cosine value for a specific angle, as shown in exercise (a) \( \cos(\pi/4) \) or exercise (b) \( \cos(5\pi/4) \) for instance, we look at the angle's terminal side when it is in standard position (originating from the positive x-axis and sweeping counter-clockwise).

An important tip is understanding that cosine values are positive in the first and fourth quadrants and negative in the second and third quadrants. This explains why \( \cos(5\pi/4) \) and \( \cos(-3\pi/4) \) from the third and second quadrants, respectively, yield negative values. Equipped with quadrant rules and knowledge about equivalent angles, determining the cosine for any given angle becomes more intuitive.
Sine Function
Similarly, the sine function plays a pivotal role in trigonometry, representing the y-coordinate of any point located on the unit circle's circumference. When dealing with the unit circle, \( \sin(\theta) \) tells us how far up or down from the x-axis this point is.

For example, when looking at exercise (d) \( \sin(5\pi/6) \) or exercise (e) \( \sin(-13\pi/6) \) we refer to the unit circle to understand that these y-coordinates are positive in the first and second quadrants and negative in the third and fourth quadrants. This is why \( \sin(5\pi/6) \) results in a positive value. Furthermore, the exercise highlights the concept of co-terminal angles, where angles that differ by full rotations (multiples of \( 2\pi \) or 360 degrees) have the same sine value. It's an essential strategy to simplify the process by reducing the angle to its acute counterpart before locating it on the unit circle.

Sine values help calculate various properties, such as the height of an object when the angle of elevation and distance are known. It is also used in wave functions to model periodic phenomena.
Trigonometric Functions
The concept of trigonometric functions extends beyond just sine and cosine, incorporating other functions like tangent, cotangent, secant, and cosecant. These functions all relate to the angles and sides of right-angled triangles and points on the unit circle.

In a broader sense, trigonometric functions are the building blocks of trigonometry, used to describe oscillatory and periodic behavior in various scientific and engineering fields. They allow us to solve problems involving triangles, model motion, understand sound and light waves, and even in designing complex structures like bridges or roller coasters.

Understanding how each trigonometric function behaves and interrelates with others within the unit circle can significantly aid in solving complex trigonometric problems. Remember that each trigonometric function has a specific domain and range, and they exhibit periodicity, which means they repeat values at regular intervals. For students exploring these functions, a solid grasp of the unit circle, angle measurement, and the behavior of trigonometric functions within the different quadrants is crucial.

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Most popular questions from this chapter

On the same set of axes sketch the graphs of \(f(x)\) and \(f^{-1}(x)\). (a) \(f(x)=\sin x\), where \(x \in[-\pi / 2, \pi / 2]\). (b) \(f(x)=\cos x\), where \(x \in[0, \pi]\).

Match with each of the expressions on the right with the equivalent expression on the left. (a) \(\cos ^{2} x\) (i) \(1-\cos ^{2} x\) (b) \(\sin ^{2} x\) (ii) \(\quad 1-\sin ^{2} x\) (c) \(\tan ^{2} x\) (iii) \(\frac{1}{2}-\frac{1}{2} \cos 2 x\) (d) \(\sin x\) (iv) \(-1+\sec ^{2} x\) (e) \(\cos x\) (v) \(\quad 1-\sin ^{2}(x+\pi / 2)\) (f) \(-\cos x\) (vi) \(\sin (x-\pi / 2)\) (vii) \(\frac{1}{2}+\frac{1}{2} \cos 2 x\) (viii) \(\cos (x-\pi / 2)\)

\text { If } x \text { is a solution to } \cos x \sin x \cos (2 x) \sin (2 x)=0, \text { find } a l l \text { possible values of } \cos (x)

We are standing on flat ground in Monument Valley trying to estimate the height of the edifices. We have surveying equipment and take all of our measurements from a height of 5 feet. We find the angle of elevation to the top of one structure is \(23^{\circ} .\) We move 500 feet closer to the structure and find that the angle of elevation is now \(29^{\circ} .\) How tall is the structure?

Force \(A\) has a horizontal component of 3 pounds and a vertical component of 4 pounds. Force \(B\) has a horizontal component of 5 pounds and a vertical component of 12 pounds. (a) What is the strength of force \(A\) ? What angle does this force vector make with the horizontal? (Give a numerical approximation in degrees.) (b) What is the strength of force \(B\) ? What angle does this force vector make with the horizontal? (Give a numerical approximation in degrees.) (c) What is the component of force \(A\) in the direction of force \(B\) ?

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