Get expert insights and reliable answers to your questions on IDNLearn.com. Discover detailed and accurate answers to your questions from our knowledgeable and dedicated community members.
Sagot :
To find the exact value of [tex]\(\tan 15^\circ\)[/tex] without using a calculator, we can use the tangent subtraction formula:
[tex]\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \][/tex]
We choose [tex]\(A = 45^\circ\)[/tex] and [tex]\(B = 30^\circ\)[/tex]. Thus,
[tex]\[ \tan(15^\circ) = \tan(45^\circ - 30^\circ) \][/tex]
We know the exact values for [tex]\(\tan 45^\circ\)[/tex] and [tex]\(\tan 30^\circ\)[/tex]:
[tex]\[ \tan 45^\circ = 1 \][/tex]
[tex]\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \][/tex]
Now, apply these values to the formula:
[tex]\[ \tan(15^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \][/tex]
[tex]\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \][/tex]
Next, let's simplify both the numerator and the denominator separately.
First, simplify the numerator:
[tex]\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \][/tex]
Now, simplify the denominator:
[tex]\[ 1 + 1 \cdot \frac{1}{\sqrt{3}} = 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}} \][/tex]
Next, we combine these simplified forms into the original formula:
[tex]\[ \tan(15^\circ) = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} \][/tex]
Since both the numerator and the denominator are fractions with a common denominator of [tex]\(\sqrt{3}\)[/tex], we can simplify this to:
[tex]\[ \tan(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \][/tex]
To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator, [tex]\(\sqrt{3} - 1\)[/tex]:
[tex]\[ \tan(15^\circ) = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} \][/tex]
[tex]\[ \tan(15^\circ) = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2} \][/tex]
Now simplify the expression further:
[tex]\[ (\sqrt{3} - 1)^2 = (\sqrt{3})^2 - 2 \cdot \sqrt{3} \cdot 1 + 1^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \][/tex]
[tex]\[ (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2 \][/tex]
Thus:
[tex]\[ \tan(15^\circ) = \frac{4 - 2\sqrt{3}}{2} = \frac{4}{2} - \frac{2\sqrt{3}}{2} = 2 - \sqrt{3} \][/tex]
Finally, simplifying further:
[tex]\[ \tan(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = 0.26794919243112264 \][/tex]
We find that the value of [tex]\(\tan 15^\circ\)[/tex] is approximately [tex]\(0.26794919243112264\)[/tex].
[tex]\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \][/tex]
We choose [tex]\(A = 45^\circ\)[/tex] and [tex]\(B = 30^\circ\)[/tex]. Thus,
[tex]\[ \tan(15^\circ) = \tan(45^\circ - 30^\circ) \][/tex]
We know the exact values for [tex]\(\tan 45^\circ\)[/tex] and [tex]\(\tan 30^\circ\)[/tex]:
[tex]\[ \tan 45^\circ = 1 \][/tex]
[tex]\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \][/tex]
Now, apply these values to the formula:
[tex]\[ \tan(15^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \][/tex]
[tex]\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \][/tex]
Next, let's simplify both the numerator and the denominator separately.
First, simplify the numerator:
[tex]\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \][/tex]
Now, simplify the denominator:
[tex]\[ 1 + 1 \cdot \frac{1}{\sqrt{3}} = 1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}} \][/tex]
Next, we combine these simplified forms into the original formula:
[tex]\[ \tan(15^\circ) = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} \][/tex]
Since both the numerator and the denominator are fractions with a common denominator of [tex]\(\sqrt{3}\)[/tex], we can simplify this to:
[tex]\[ \tan(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \][/tex]
To rationalize the denominator, multiply both the numerator and the denominator by the conjugate of the denominator, [tex]\(\sqrt{3} - 1\)[/tex]:
[tex]\[ \tan(15^\circ) = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} \][/tex]
[tex]\[ \tan(15^\circ) = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2} \][/tex]
Now simplify the expression further:
[tex]\[ (\sqrt{3} - 1)^2 = (\sqrt{3})^2 - 2 \cdot \sqrt{3} \cdot 1 + 1^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} \][/tex]
[tex]\[ (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2 \][/tex]
Thus:
[tex]\[ \tan(15^\circ) = \frac{4 - 2\sqrt{3}}{2} = \frac{4}{2} - \frac{2\sqrt{3}}{2} = 2 - \sqrt{3} \][/tex]
Finally, simplifying further:
[tex]\[ \tan(15^\circ) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = 0.26794919243112264 \][/tex]
We find that the value of [tex]\(\tan 15^\circ\)[/tex] is approximately [tex]\(0.26794919243112264\)[/tex].
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.
