Problem 31
Question
In Exercises \(25-32,\) the unit circle has been divided into eight equal arcs, corresponding to t-values of $$ 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \pi, \frac{5 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}, \text { and } 2 \pi $$ a. Use the \((x, y)\) coordinates in the figure to find the value of the trigonometric function. b. Use periodic properties and your answer from part (a) to find the value of the same trigonometric function at the indicated real number. $$ \begin{aligned} &a.\sin \frac{7 \pi}{4}\\\ &b.\sin \frac{47 \pi}{4} \end{aligned} $$
Step-by-Step Solution
Verified Answer
a. \(sin(\frac{7 \pi}{4}) = -\sqrt{2}/2\), b. \(sin(\frac{47 \pi}{4}) = -\sqrt{2}/2\).
1Step 1: Finding the \(sin(\frac{7 \pi}{4})\)
The unit circle is divided into 8 equal parts, each corresponding to an angle and specific t-values. Note that the angle \(\frac{7 \pi}{4}\) is the same angle as \(-\frac{\pi}{4}\), both of which lie in the fourth quadrant where sine is negative. Also, in a unit circle, the sine function represents the y-coordinate. For an angle of \(\pi/4\) or \(7\pi/4\), the y-coordinate is \(\ sqrt{2}/2\). As sine is negative in the fourth quadrant, the \(sin(\frac{7 \pi}{4})\) is \(-\sqrt{2}/2\).
2Step 2: Using the Periodic Property of Sine to find \(sin(\frac{47 \pi}{4})\)
The sine function is periodic with a period of \(2 \pi\), which means that the \(\sin(\frac{47 \pi}{4})\) is the same as \(\sin(\frac{7 \pi}{4})\). So, \(sin(\frac{47 \pi}{4}) = -\sqrt{2}/2\).
Key Concepts
Sine FunctionPeriodic Properties of Trigonometric FunctionsRadians and Degrees Conversion
Sine Function
Imagine standing at the center of a clock face, and as the hand sweeps around, you're only looking at how high or low it points. That's similar to how the sine function works with circles.
The sine function, often abbreviated as 'sin', relates an angle in a right triangle to the ratio of the length of the side opposite to the angle and the hypotenuse, which is the longest side. But when we move onto the unit circle, we keep things even simpler. The unit circle has a radius of 1, and for any angle you choose, sin will tell you the height of the point where the radius meets the circle. This 'height' is actually the y-coordinate of the point on the circle's edge.
For example, in the unit circle, as you discovered in the exercise, the sine of \( \frac{7\pi}{4} \) is \( -\sqrt{2}/2 \) because it's the y-coordinate at that angle, and the point is located below the horizontal diameter, thus the negative sign.
The sine function, often abbreviated as 'sin', relates an angle in a right triangle to the ratio of the length of the side opposite to the angle and the hypotenuse, which is the longest side. But when we move onto the unit circle, we keep things even simpler. The unit circle has a radius of 1, and for any angle you choose, sin will tell you the height of the point where the radius meets the circle. This 'height' is actually the y-coordinate of the point on the circle's edge.
For example, in the unit circle, as you discovered in the exercise, the sine of \( \frac{7\pi}{4} \) is \( -\sqrt{2}/2 \) because it's the y-coordinate at that angle, and the point is located below the horizontal diameter, thus the negative sign.
Periodic Properties of Trigonometric Functions
Unlike rides at an amusement park, trig functions like sine don't have a start or end; they keep going around in circles forever. That's because they have a 'period'.
The period of a function is the length it needs to start repeating the same values. For the sine function, that period is \(2\pi\) radians, which means if you walk around our clock face of the unit circle, once you're back where you started (after \(2\pi\) radians), the sine will start over, just like the 12 hours on a clock.
This is why, in the exercise, sine didn't care whether you were looking at \( \frac{7\pi}{4} \) or \( \frac{47\pi}{4} \). Both angles ultimately point to the same spot on the unit circle, because you’ve gone around whole extra turns, but the height (the y-coordinate) at which they end up is precisely the same.
The period of a function is the length it needs to start repeating the same values. For the sine function, that period is \(2\pi\) radians, which means if you walk around our clock face of the unit circle, once you're back where you started (after \(2\pi\) radians), the sine will start over, just like the 12 hours on a clock.
This is why, in the exercise, sine didn't care whether you were looking at \( \frac{7\pi}{4} \) or \( \frac{47\pi}{4} \). Both angles ultimately point to the same spot on the unit circle, because you’ve gone around whole extra turns, but the height (the y-coordinate) at which they end up is precisely the same.
Radians and Degrees Conversion
Angles can be measured using two main units: degrees and radians. While most people are familiar with degrees because of protractors and compass directions, radians are a bit like the secret language of mathematicians.
A full circle is 360 degrees, right? In radian language, that's \(2\pi\) radians. Why pi? Because radian measures are based on the circle's circumference, which is related to \(\pi\). One radian is roughly 57.3 degrees. To switch between the two, you can use these handy formulas:
A full circle is 360 degrees, right? In radian language, that's \(2\pi\) radians. Why pi? Because radian measures are based on the circle's circumference, which is related to \(\pi\). One radian is roughly 57.3 degrees. To switch between the two, you can use these handy formulas:
- To convert degrees to radians, multiply by \(\pi/180\).
- To convert radians to degrees, multiply by \(180/\pi\).
Other exercises in this chapter
Problem 30
In Exercises \(29-34,\) convert each angle in degrees to radians. Round to two decimal places. $$ 76^{\circ} $$
View solution Problem 31
Find the exact value of each expression, if possible. Do not use a calculator. $$ \sin \left(\sin ^{-1} 0.9\right) $$
View solution Problem 31
Determine the amplitude of each function. Then graph the function and \(y=\cos x\) in the same rectangular coordinate system for \(0 \leq x \leq 2 \pi\). $$y=2
View solution Problem 31
In Exercises 29–44, graph two periods of the given cosecant or secant function. $$ y=\frac{1}{2} \csc \frac{x}{2} $$
View solution