Problem 80
Question
(a) Show that \( \cos (x^2) \ge \cos x \) for \( 0 \le x \le 1 \). (b) Deduce that \( \displaystyle \int^{\pi/6}_0 \cos (x^2) \, dx \ge \frac{1}{2} \).
Step-by-Step Solution
Verified Answer
(a) Use the fact that \( \cos \) decreases and \( x^2 \le x \). (b) Integrals comparison yields \( \int^{\pi/6}_0 \cos(x^2) \, dx \ge \frac{1}{2} \).
1Step 1: Understand the Function Behavior
To show that \( \cos (x^2) \ge \cos x \) for \( 0 \le x \le 1 \), we first need to understand the function behavior. Since \( \cos x \) is a decreasing function for \( 0 \le x \le \pi \), and \( x^2 \le x \) for \( 0 \le x \le 1 \), we can infer that \( \cos (x^2) \) will tend to be greater than \( \cos x \) within this interval, as a smaller angle in cosine means a larger value.
2Step 2: Verify the Trigonometric Inequality
We compare the values of \( \cos(x^2) \) and \( \cos(x) \) directly for a few values in the given range to verify the inequality. For example, for \( x = 0\) and \( x = 1 \), we have \( \cos(0^2) = \cos(0) = 1 \) and \( \cos(1^2) = \cos(1) \approx 0.5403 \). Since \( \cos \) is a decreasing function, it supports the inequality that \( \cos (x^2) \ge \cos x \) in the interval.
3Step 3: Deduce the Integral Inequality
To deduce that \( \int^{\pi/6}_0 \cos(x^2) \, dx \ge \frac{1}{2} \), start by acknowledging that \( \cos(x^2) \ge \cos(x) \). Since we know that \( \int^{\pi/6}_0 \cos x \, dx \) evaluates to \( \sin(\pi/6) - \sin(0) = \frac{1}{2} \), and \( \cos(x^2) \ge \cos(x) \), it follows that \( \int^{\pi/6}_0 \cos(x^2) \, dx \ge \frac{1}{2} \).
4Step 4: Conclude with the Comparison Outcome
By comparing both integral values, we conclude that since \( \cos(x^2) \ge \cos(x) \) throughout \([0,\pi/6]\), it effectively means the area under \( \cos(x^2) \) in this interval is greater than or equal to the area under \( \cos(x) \). Hence, \( \int^{\pi/6}_0 \cos(x^2) \, dx \ge 1/2 \).
Key Concepts
Trigonometric FunctionsIntegral CalculusCosine Function
Trigonometric Functions
Trigonometric functions are fundamental mathematical functions that relate the angles of a triangle to the lengths of its sides. These are significant in various fields such as physics, engineering, and mathematics. The primary trigonometric functions include sine, cosine, tangent, and their reciprocals: cosecant, secant, and cotangent.
In particular, the cosine function, which is central to our discussion, is defined for an angle in a right-angled triangle as the ratio of the length of the adjacent side to the hypotenuse. Mathematically, it is expressed as:\[ ext{cos}(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\]The cosine function has some key properties:
In particular, the cosine function, which is central to our discussion, is defined for an angle in a right-angled triangle as the ratio of the length of the adjacent side to the hypotenuse. Mathematically, it is expressed as:\[ ext{cos}(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\]The cosine function has some key properties:
- It is periodic with a period of \(2\pi\).
- Its range is from -1 to 1.
- It is even, meaning \(\cos(-x) = \cos(x)\).
- It decreases in the interval \(0 \le x \le \pi\).
Integral Calculus
Integral calculus is concerned with the accumulation of quantities and the areas under curves. Unlike differential calculus, which deals with the rate of change, integral calculus focuses on how much there is.
One of the fundamental concepts in integral calculus is the definite integral, which is used to compute the area under a curve defined by a function \(f(x)\) over an interval \([a, b]\). It is expressed as:\[\int_a^b f(x) \, dx\]This provides the net area between the function and the x-axis from \(x = a\) to \(x = b\). In problems involving trigonometric functions, such as finding the area under a cosine function, integral calculus becomes essential.
In the given solution, we calculate the integral of \(\cos(x^2)\) from 0 to \(\frac{\pi}{6}\) and compare it to the integral of \(\cos(x)\), demonstrating that trigonometric inequalities can be expressed and solved through integral calculus by comparing areas.
One of the fundamental concepts in integral calculus is the definite integral, which is used to compute the area under a curve defined by a function \(f(x)\) over an interval \([a, b]\). It is expressed as:\[\int_a^b f(x) \, dx\]This provides the net area between the function and the x-axis from \(x = a\) to \(x = b\). In problems involving trigonometric functions, such as finding the area under a cosine function, integral calculus becomes essential.
In the given solution, we calculate the integral of \(\cos(x^2)\) from 0 to \(\frac{\pi}{6}\) and compare it to the integral of \(\cos(x)\), demonstrating that trigonometric inequalities can be expressed and solved through integral calculus by comparing areas.
Cosine Function
The cosine function is a trigonometric function that is very familiar in geometry and calculus. It oscillates between the values of -1 to 1, offering a periodic pattern that repeats every \(2\pi\). This makes it extremely useful for modeling wave-like patterns in data.
An important characteristic of the cosine function, which is exploited in the problem, is that it is decreasing on the interval \(0 \le x \le \pi\). This behavior is why in the solution, \(\cos(x^2)\) can be shown to be larger than \(\cos(x)\) when \(x^2\) is less than \(x\), and both angles are in the increasing zone of the function.
The importance of this property can be seen in practical applications. For instance:
An important characteristic of the cosine function, which is exploited in the problem, is that it is decreasing on the interval \(0 \le x \le \pi\). This behavior is why in the solution, \(\cos(x^2)\) can be shown to be larger than \(\cos(x)\) when \(x^2\) is less than \(x\), and both angles are in the increasing zone of the function.
The importance of this property can be seen in practical applications. For instance:
- In physics, it helps in determining the phase differences in oscillating systems.
- In engineering, cosine is used in signal processing to create waveforms.
Other exercises in this chapter
Problem 78
If \( f \) is continuous and \( g \) and \( h \) are differentiable functions, find a formula for $$ \displaystyle \frac{d}{dx} \int^{h(x)}_{g(x)} f(t) \, dt $$
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A model for the basal metabolism rate, in kcal/h, of a young man is \( R(t) = 85 - 0.18 \cos(\pi t/12) \), where \( t \) is the time in hours measured from 5:00
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An oil storage tank ruptures at time \( t = 0 \) and oil leaks from the tank at a rate of \( r(t) = 100e^{-0.01t} \) liters per minute. How much oil leaks out d
View solution Problem 81
Show that $$ 0 \le \int^{10}_5 \frac{x^2}{x^4 + x^2 + 1} \, dx \le 0.1 $$ by comparing the integrand to a simpler function.
View solution