Chapter 10: Problem 3
Solve each equation over the interval \([0,2 \pi)\) $$\sin \frac{x}{2}=\cos \frac{x}{2}$$
Short Answer
Expert verified
\(x = \frac{\pi}{2}\) is the only solution in the interval \([0, 2\pi)\).
Step by step solution
01
Understand the problem statement
We need to solve the equation \(\sin \frac{x}{2} = \cos \frac{x}{2}\) within the interval \([0, 2\pi)\). This interval indicates the values of \(x\) that start from 0 and go up to, but do not include, \(2\pi\).
02
Transform the equation
Recognize that \(\sin \theta = \cos \theta\) implies \(\theta = \frac{\pi}{4} + n\pi\) for some integer \(n\). This is because the sine and cosine functions are equal at these points.
03
Apply the transformation to the equation
Substitute \(\theta = \frac{x}{2}\) into \(\theta = \frac{\pi}{4} + n\pi\) to get \(\frac{x}{2} = \frac{\pi}{4} + n\pi\).
04
Solve for x
Multiply both sides of the equation \(\frac{x}{2} = \frac{\pi}{4} + n\pi\) by 2 to solve for \(x\): \(x = \frac{\pi}{2} + 2n\pi\).
05
Find valid solutions in the interval
Determine the values of \(n\) such that \(0 \leq x < 2\pi\). - For \(n = 0\), \(x = \frac{\pi}{2}\).- For \(n = 1\), \(x = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}\) (not valid since \(\frac{5\pi}{2} > 2\pi\)). - For \(n = -1\), \(x = \frac{\pi}{2} - 2\pi = -\frac{3\pi}{2}\) (not valid since negative).
06
Confirm the solutions
Double-check each solution for validity within the specified interval. The only valid value for \(x\) within \([0, 2\pi)\) from the calculations above is \(x = \frac{\pi}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interval Notation
Interval notation is a mathematical shorthand used to describe the set of all numbers between a pair of given numbers. It gives a clear way to represent which numbers belong to a set. For example, the interval \([0, 2\pi)\) includes all numbers starting from 0 up to, but not including, \(2\pi\). The bracket \("["\) indicates that the number is included in the interval while the parenthesis \(")"\) shows that the number is not included.
- \([a, b]\) means all numbers \(x\) where \(a \leq x \leq b\).
- \([a, b)\) means all numbers \(x\) where \(a \leq x < b\).
- \((a, b]\) means all numbers \(x\) where \(a < x \leq b\).
- \((a, b)\) means all numbers \(x\) where \(a < x < b\).
Sine Function
The sine function is one of the fundamental functions in trigonometry. It relates the angle of a right-angled triangle to the ratio of the length of the opposite side over the hypotenuse. In a unit circle, which is a circle with a radius of 1, the sine of an angle \(\theta\) can be understood as the y-coordinate of the point where the terminal side of the angle intersects the circle.Here are some key properties of the sine function:
- It is periodic, repeating every \(2\pi\) radians.
- The range of sine is \([-1, 1]\), meaning that it will never exceed these values.
- It has specific symmetry properties, such as being an odd function, satisfying \(\sin(-\theta) = -\sin(\theta)\).
Cosine Function
The cosine function is equally crucial in trigonometry, similar to the sine function. It defines the ratio of the length of the adjacent side to the hypotenuse in a right triangle. Like sine, in the context of a unit circle, the cosine of an angle \(\theta\) is the x-coordinate of the point of intersection with the circle.Important aspects of the cosine function include:
- Its periodic nature, repeating every \(2\pi\) radians.
- The range of values is also constrained between \([-1, 1]\).
- It is an even function, meaning \(\cos(-\theta) = \cos(\theta)\), which shows its symmetric property along the y-axis.