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Magazine Chapter 7: Problem 62
Find the following. \(\sin \left(2 \cos ^{-1} \frac{3}{5}\right)\)
Short Answer
Expert verified
\(\frac{24}{25}\)
Step by step solution
01
- Understand the Problem
The goal is to find the value of \(\text{\sin} \left(2 \text{\cos}^{-1} \frac{3}{5} \right)\). We need to use trigonometric identities to simplify this expression.
02
- Use the Inverse Cosine Function
Let \(\theta = \text{\cos}^{-1} \frac{3}{5}\). By the definition of the inverse cosine function, \(\cos \theta = \frac{3}{5}\).
03
- Find \sin \theta
Since \(\theta\) is an angle in a right-angled triangle, use the Pythagorean theorem to find \(\text{\sin} \theta\). \[ \text{\sin}^2 \theta + \text{\cos}^2 \theta = 1 \]. Substituting \(\text{\cos} \theta = \frac{3}{5}\), we get: \[ \text{\sin}^2 \theta + \left( \frac{3}{5} \right)^2 = 1 \] \[ \text{\sin}^2 \theta + \frac{9}{25} = 1 \] \[ \text{\sin}^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \] \[ \text{\sin} \theta = \frac{4}{5} \]
04
- Use the Double-Angle Formula
To find \(\text{\sin} (2\theta)\), use the double-angle formula for sine: \[ \text{\sin}(2\theta) = 2 \text{\sin} \theta \text{\cos} \theta \]. Substitute the values found: \[ \text{\sin}(2\theta) = 2 \left( \frac{4}{5} \right) \left( \frac{3}{5} \right) \]
05
- Simplify the Expression
Calculate the values: \[ 2 \left( \frac{4}{5} \right) \left( \frac{3}{5} \right) = 2 \times \frac{12}{25} = \frac{24}{25} \]. Therefore, \(\text{\sin} (2 \text{\cos}^{-1} \frac{3}{5}) = \frac{24}{25}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Trigonometric Functions
Inverse trigonometric functions allow us to determine angles when given a trigonometric ratio. For example, \(\text{cos}^{-1}\) or \(\text{acos}\) is used to find an angle when the cosine of that angle is known.
Let鈥檚 consider \(\theta\) such that \(\theta = \text{cos}^{-1} \frac{3}{5}\). This means \(\text{cos}(\theta) = \frac{3}{5}\). The angle \(\theta\) corresponds to the angle in a right-angled triangle where the adjacent side (against the angle) is 3 units and the hypotenuse is 5 units.
Inverse functions are crucial because they help us work backwards from ratios to angles, which we can then use in more complex trigonometric calculations, such as those involving double-angle formulas.
Let鈥檚 consider \(\theta\) such that \(\theta = \text{cos}^{-1} \frac{3}{5}\). This means \(\text{cos}(\theta) = \frac{3}{5}\). The angle \(\theta\) corresponds to the angle in a right-angled triangle where the adjacent side (against the angle) is 3 units and the hypotenuse is 5 units.
Inverse functions are crucial because they help us work backwards from ratios to angles, which we can then use in more complex trigonometric calculations, such as those involving double-angle formulas.
Pythagorean Theorem
The Pythagorean theorem is an essential concept in geometry and trigonometry. It states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Mathematically: \[a^2 + b^2 = c^2\] In our exercise, since we know \(\text{cos}(\theta) = \frac{3}{5}\), we can use the Pythagorean theorem to find \(\text{sin}(\theta)\).
We know: \(\text{cos}(\theta) = \frac{3}{5}\), thus \[\text{cos}^2(\theta) = \frac{9}{25}\]
Applying the theorem:
\[\text{sin}^2(\theta) + \text{cos}^2(\theta) = 1\]
\[\text{sin}^2(\theta) + \frac{9}{25} = 1\]
\[\text{sin}^2(\theta) = 1 - \frac{9}{25}\]
\[\text{sin}^2(\theta) = \frac{16}{25}\]
Thus, \(\text{sin}(\theta) = \frac{4}{5}\).
This shows how Pythagorean theorem helps in finding unknown trigonometric ratios.
Mathematically: \[a^2 + b^2 = c^2\] In our exercise, since we know \(\text{cos}(\theta) = \frac{3}{5}\), we can use the Pythagorean theorem to find \(\text{sin}(\theta)\).
We know: \(\text{cos}(\theta) = \frac{3}{5}\), thus \[\text{cos}^2(\theta) = \frac{9}{25}\]
Applying the theorem:
\[\text{sin}^2(\theta) + \text{cos}^2(\theta) = 1\]
\[\text{sin}^2(\theta) + \frac{9}{25} = 1\]
\[\text{sin}^2(\theta) = 1 - \frac{9}{25}\]
\[\text{sin}^2(\theta) = \frac{16}{25}\]
Thus, \(\text{sin}(\theta) = \frac{4}{5}\).
This shows how Pythagorean theorem helps in finding unknown trigonometric ratios.
Double-Angle Formula
The double-angle formulas are used to express trigonometric functions of double angles (2胃) in terms of single angles (胃). For sine, the double-angle formula is:
\[\text{sin}(2\theta) = 2 \text{sin}(\theta) \text{cos}(\theta)\].
In our problem, we need to find \(\text{sin}\big(2 \text{cos}^{-1} \frac{3}{5}\big)\). We already determined that \(\text{cos}(\theta) = \frac{3}{5}\) and \(\text{sin}(\theta) = \frac{4}{5}\). Using the double-angle formula for sine:
\[\text{sin}(2\theta) = 2 \big(\frac{4}{5}\big) \big(\frac{3}{5}\big)\]
\[ \text{sin}(2\theta) = 2 \big(\frac{12}{25}\big)\]
\[ \text{sin}(2\theta) = \frac{24}{25}\].
This result helps us see how double-angle formulas simplify the computation of trigonometric values for double angles.
\[\text{sin}(2\theta) = 2 \text{sin}(\theta) \text{cos}(\theta)\].
In our problem, we need to find \(\text{sin}\big(2 \text{cos}^{-1} \frac{3}{5}\big)\). We already determined that \(\text{cos}(\theta) = \frac{3}{5}\) and \(\text{sin}(\theta) = \frac{4}{5}\). Using the double-angle formula for sine:
\[\text{sin}(2\theta) = 2 \big(\frac{4}{5}\big) \big(\frac{3}{5}\big)\]
\[ \text{sin}(2\theta) = 2 \big(\frac{12}{25}\big)\]
\[ \text{sin}(2\theta) = \frac{24}{25}\].
This result helps us see how double-angle formulas simplify the computation of trigonometric values for double angles.
