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Chapter 5: Problem 30

Find the smallest positive number \(\theta\) such that \(10^{\sin \theta}=7\).

Short Answer

Expert verified
The smallest positive number \(\theta\) such that \(10^{\sin \theta} = 7\) is approximately \(\theta \approx 1.017\, \text{radians}\).

Step by step solution

01

Rewrite the equation in logarithmic form

Using the logarithm properties, we can rewrite the given equation \(10^{\sin \theta}=7\) in logarithmic form. This will give us: \(\sin \theta = \log_{10}7\)
02

Isolate 胃 by using inverse sin function

To isolate \(\theta\), we will take the inverse sine (also known as arcsin or sin^{-1}) of both sides of the equation: \(\theta = \arcsin{(\log_{10}7)}\)
03

Calculate 胃

Now, we can calculate the value of \(\theta\) using a calculator: \(\theta \approx \arcsin{(0.8451)} \approx 1.017\, \text{radians}\)
04

Ensure the smallest positive value of 胃

As trigonometric functions are periodic, for sine the period is \(2\pi\), meaning that for any positive integer k, \(\sin{\theta} = \sin{(\theta+k2\pi)}\). In this case, we see that \(\theta \approx 1.017\, \text{radians}\) is the smallest positive number that fulfills the condition, as adding any multiples of \(2\pi\) would result in a larger value. Therefore, the smallest positive number \(\theta\) such that \(10^{\sin \theta} = 7\) is \(\theta \approx 1.017\, \text{radians}\).

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Key Concepts

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

Logarithmic form
The term "logarithmic form" refers to expressing exponential equations in the form of a logarithm. This is a powerful tool when dealing with equations that involve exponents. For example, the equation \( 10^{\sin \theta} = 7 \) can be transformed into a logarithmic form to help simplify and solve it.
  • The base of the logarithm corresponds to the base of the exponent. In our example, it鈥檚 10.
  • The log result (logarithmic value) will equal the power that the base is raised to, which is \( \sin \theta \).
  • Thus, the logarithmic form becomes \( \log_{10} 7 = \sin \theta \).
Adopting the logarithmic form makes solving such equations more straightforward. It transforms complex exponential problems into simpler arithmetic operations. By understanding this principle, students can tackle a variety of similar mathematical challenges with greater confidence.
Inverse trigonometric functions
Inverse trigonometric functions allow you to determine angles given a certain trigonometric ratio. Common inverse trigonometric functions include arcsin, arccos, and arctan. They are particularly useful in converting trigonometric values back to their angles.
  • By using \( \arcsin \), you can find an angle when you know its sine value.
  • In the example solution, we use \( \theta = \arcsin{(\log_{10}7)} \).
  • This equates \( \theta \) to the angle whose sine is \( \log_{10} 7 \).
Understanding inverse functions is crucial because they decode situations where the reverse of usual trigonometric calculations is needed. This concept applies widely in fields like engineering and physics, where precise angle measurements are necessary based on known ratios.
Radians
Radians provide a convenient way to measure angles based on the radius of a circle. Unlike degrees, which divide a circle into 360 parts, radians are based on the circle's geometry, where one complete revolution equals \(2\pi\) radians. Hence, radians offer a natural and direct way to express angles.
  • A half-circle, or 180 degrees, equals \(\pi\) radians.
  • A quarter circle, or 90 degrees, is \(\frac{\pi}{2}\) radians.
  • Radians are often used in mathematical contexts due to their precision and mathematical simplicity.
In the provided solution, the value of \( \theta \) is calculated to be approximately \(1.017\) radians. Recognizing the concept of radians is essential for understanding trigonometric functions and their periodic nature, which is especially important when solving equations like the one in the exercise.

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Most popular questions from this chapter

Evaluate the indicated expressions assuming that $$ \begin{array}{ll} \cos x=\frac{1}{3} & \text { and } \sin y=\frac{1}{4} \\ \sin u=\frac{2}{3} & \text { and } \cos v=\frac{1}{5} \end{array} $$ Assume also that \(x\) and \(u\) are in the interval \(\left(0, \frac{\pi}{2}\right),\) that \(y\) is in the interval \(\left(\frac{\pi}{2}, \pi\right),\) and that \(v\) is in the interval \(\left(-\frac{\pi}{2}, 0\right) .\) $$\sin (u+v)$$

Evaluate the indicated expressions assuming that $$ \begin{array}{ll} \cos x=\frac{1}{3} & \text { and } \sin y=\frac{1}{4} \\ \sin u=\frac{2}{3} & \text { and } \cos v=\frac{1}{5} \end{array} $$ Assume also that \(x\) and \(u\) are in the interval \(\left(0, \frac{\pi}{2}\right),\) that \(y\) is in the interval \(\left(\frac{\pi}{2}, \pi\right),\) and that \(v\) is in the interval \(\left(-\frac{\pi}{2}, 0\right) .\) $$\cos (x-y)$$

Evaluate the given quantities assuming that \(u\) and \(v\) are both in the interval \(\left(-\frac{\pi}{2}, 0\right)\) and \(\tan u=-\frac{1}{7} \quad\) and \(\quad \tan v=-\frac{1}{8}\) $$\sin u$$

Find a formula for \(\tan \left(\theta-\frac{\pi}{4}\right)\).

Suppose \(\theta\) is not an integer multiple of \(\pi\). Explain why the point \((1,2 \cos \theta)\) is on the line containing the point \((\sin \theta, \sin (2 \theta))\) and the origin.
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