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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].
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