Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To determine which option represents the exact value of [tex]\(\tan \left(\frac{\pi}{12}\right)\)[/tex], we start by understanding the tangent value for the angle [tex]\(\frac{\pi}{12}\)[/tex].
The angle [tex]\(\frac{\pi}{12}\)[/tex] radians can be expressed in degrees as [tex]\(15^\circ\)[/tex] since [tex]\(\frac{180^\circ}{12} = 15^\circ\)[/tex].
Using known trigonometric identities, one approach to find [tex]\(\tan 15^\circ\)[/tex] is recognizing that [tex]\(15^\circ\)[/tex] can be represented as the difference between two familiar angles: [tex]\(45^\circ\)[/tex] and [tex]\(30^\circ\)[/tex]. Therefore, we can use the tangent subtraction formula:
[tex]\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \][/tex]
For [tex]\(a = 45^\circ\)[/tex] and [tex]\(b = 30^\circ\)[/tex]:
[tex]\[ \tan 15^\circ = \tan (45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \][/tex]
We know the tangent values:
[tex]\[ \tan 45^\circ = 1 \quad \text{and} \quad \tan 30^\circ = \frac{1}{\sqrt{3}} \][/tex]
Substituting these into the formula gives us:
[tex]\[ \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \][/tex]
To simplify the fraction, we'll first combine the terms in the numerator and denominator to have a common denominator of [tex]\(\sqrt{3}\)[/tex]:
[tex]\[ \tan 15^\circ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \][/tex]
Next, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator [tex]\(\sqrt{3} - 1\)[/tex]:
[tex]\[ \tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} \][/tex]
[tex]\[ \tan 15^\circ = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \][/tex]
Thus, we have:
[tex]\[ \tan \left( \frac{\pi}{12} \right) = 2 - \sqrt{3} \][/tex]
We now compare this result with the given options:
1. [tex]\(\frac{1}{2 + \sqrt{3}}\)[/tex]
2. [tex]\(\frac{1}{\sqrt{3}}\)[/tex]
3. [tex]\(\frac{3}{\sqrt{3}}\)[/tex]
4. [tex]\(\frac{1}{2 - \sqrt{3}}\)[/tex]
Evaluating the first option:
[tex]\[ \frac{1}{2 + \sqrt{3}} \][/tex]
We need to rationalize the denominator by multiplying by [tex]\(2 - \sqrt{3}\)[/tex]:
[tex]\[ \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \][/tex]
Now, we see that:
[tex]\[ \tan \left(\frac{\pi}{12}\right) = 2 - \sqrt{3} = \frac{1}{2 + \sqrt{3}} \][/tex]
Thus, the correct answer is the first option:
[tex]\[ \boxed{\frac{1}{2 + \sqrt{3}}} \][/tex]
The angle [tex]\(\frac{\pi}{12}\)[/tex] radians can be expressed in degrees as [tex]\(15^\circ\)[/tex] since [tex]\(\frac{180^\circ}{12} = 15^\circ\)[/tex].
Using known trigonometric identities, one approach to find [tex]\(\tan 15^\circ\)[/tex] is recognizing that [tex]\(15^\circ\)[/tex] can be represented as the difference between two familiar angles: [tex]\(45^\circ\)[/tex] and [tex]\(30^\circ\)[/tex]. Therefore, we can use the tangent subtraction formula:
[tex]\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \][/tex]
For [tex]\(a = 45^\circ\)[/tex] and [tex]\(b = 30^\circ\)[/tex]:
[tex]\[ \tan 15^\circ = \tan (45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \][/tex]
We know the tangent values:
[tex]\[ \tan 45^\circ = 1 \quad \text{and} \quad \tan 30^\circ = \frac{1}{\sqrt{3}} \][/tex]
Substituting these into the formula gives us:
[tex]\[ \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \][/tex]
To simplify the fraction, we'll first combine the terms in the numerator and denominator to have a common denominator of [tex]\(\sqrt{3}\)[/tex]:
[tex]\[ \tan 15^\circ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \][/tex]
Next, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator [tex]\(\sqrt{3} - 1\)[/tex]:
[tex]\[ \tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} \][/tex]
[tex]\[ \tan 15^\circ = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \][/tex]
Thus, we have:
[tex]\[ \tan \left( \frac{\pi}{12} \right) = 2 - \sqrt{3} \][/tex]
We now compare this result with the given options:
1. [tex]\(\frac{1}{2 + \sqrt{3}}\)[/tex]
2. [tex]\(\frac{1}{\sqrt{3}}\)[/tex]
3. [tex]\(\frac{3}{\sqrt{3}}\)[/tex]
4. [tex]\(\frac{1}{2 - \sqrt{3}}\)[/tex]
Evaluating the first option:
[tex]\[ \frac{1}{2 + \sqrt{3}} \][/tex]
We need to rationalize the denominator by multiplying by [tex]\(2 - \sqrt{3}\)[/tex]:
[tex]\[ \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \][/tex]
Now, we see that:
[tex]\[ \tan \left(\frac{\pi}{12}\right) = 2 - \sqrt{3} = \frac{1}{2 + \sqrt{3}} \][/tex]
Thus, the correct answer is the first option:
[tex]\[ \boxed{\frac{1}{2 + \sqrt{3}}} \][/tex]
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
