Question

Find the value of \[\tan \dfrac{\pi }{8}\].

Answer 1

Hint: Here, we need to find the value of \[\tan \dfrac{\pi }{8}\]. We will use the formula for tangent of a double angle to form a quadratic equation in terms of \[\tan \dfrac{\pi }{8}\]. Then, using the quadratic formula, we will find the value of \[\tan \dfrac{\pi }{8}\].

Formula Used: The tangent of a double angle is given by the formula \[\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}\].
The quadratic formula states that the roots of a quadratic equation \[a{x^2} + bx + c = 0\] are given by \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\], where \[D\] is the discriminant given by the formula \[D = {b^2} - 4ac\].

Complete step-by-step answer:
We will use the formula for tangent of a double angle to find the value of \[\tan \dfrac{\pi }{8}\].
The tangent of a double angle is given by the formula \[\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}\].
Substituting \[A = \dfrac{\pi }{8}\] in the formula, we get
\[ \Rightarrow \tan \left( {2 \times \dfrac{\pi }{8}} \right) = \dfrac{{2\tan \dfrac{\pi }{8}}}{{1 - {{\tan }^2}\dfrac{\pi }{8}}}\]
Multiplying the terms in the equation, we get
\[ \Rightarrow \tan \dfrac{\pi }{4} = \dfrac{{2\tan \dfrac{\pi }{8}}}{{1 - {{\tan }^2}\dfrac{\pi }{8}}}\]
We know that the tangent of the angle \[\dfrac{\pi }{4}\] is equal to 1.
Thus, substituting \[\tan \dfrac{\pi }{4} = 1\] in the equation, we get
\[ \Rightarrow 1 = \dfrac{{2\tan \dfrac{\pi }{8}}}{{1 - {{\tan }^2}\dfrac{\pi }{8}}}\]
Simplifying the equation, we get
\[ \Rightarrow 1 - {\tan ^2}\dfrac{\pi }{8} = 2\tan \dfrac{\pi }{8}\]
Rewriting the equation, we get
\[ \Rightarrow {\tan ^2}\dfrac{\pi }{8} + 2\tan \dfrac{\pi }{8} - 1 = 0\]
Now, let \[x = \tan \dfrac{\pi }{8}\].
Therefore, the equation becomes
\[ \Rightarrow {x^2} + 2x - 1 = 0\]
This is a quadratic equation.
We will use the quadratic formula to find the roots of the quadratic equation.
The quadratic formula states that the roots of a quadratic equation \[a{x^2} + bx + c = 0\] are given by \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\], where \[D\] is the discriminant given by the formula \[D = {b^2} - 4ac\].
First, let us find the value of the discriminant.
Comparing the equation \[{x^2} + 2x - 1 = 0\] with the standard form of a quadratic equation \[a{x^2} + bx + c = 0\], we get
\[a = 1\], \[b = 2\], and \[c = - 1\]
Substituting \[a = 1\], \[b = 2\], and \[c = - 1\] in the formula for discriminant, we get
\[ \Rightarrow D = {2^2} - 4\left( 1 \right)\left( { - 1} \right)\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow D = 4 + 4\\ \Rightarrow D = 8\end{array}\]
Now, substituting \[a = 1\], \[b = 2\], and \[D = 8\] in the quadratic formula, we get
\[ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt 8 }}{{2 \times 1}}\]
Simplifying the expression, we get
\[ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 \times 2} }}{2}\]
Taking 4 out of the square root, we get
\[ \Rightarrow x = \dfrac{{ - 2 \pm 2\sqrt 2 }}{2}\]
Factoring out 2 from the numerator and simplifying, we get
\[\begin{array}{l} \Rightarrow x = \dfrac{{2\left( { - 1 \pm \sqrt 2 } \right)}}{2}\\ \Rightarrow x = - 1 \pm \sqrt 2 \end{array}\]
Therefore, either \[x = - 1 + \sqrt 2 \] or \[x = - 1 - \sqrt 2 \].
Substituting \[x = \tan \dfrac{\pi }{8}\] in the expressions, we get
\[ \Rightarrow \tan \dfrac{\pi }{8} = - 1 + \sqrt 2 \] or \[\tan \dfrac{\pi }{8} = - 1 - \sqrt 2 \]
We know that the angle \[\dfrac{\pi }{8}\] lies in the first quadrant.
Also, the tangent of any angle in the first quadrant is positive.
Therefore, the tangent of \[\dfrac{\pi }{8}\] cannot be equal to \[ - 1 - \sqrt 2 \].
Thus, we get
\[\tan \dfrac{\pi }{8} = - 1 + \sqrt 2 \]
\[\therefore\] The value of \[\tan \dfrac{\pi }{8}\] is \[ - 1 + \sqrt 2 \].

Note: We can also solve this equation by first finding the sine and cosine of the angle \[\dfrac{\pi }{8}\], and then dividing them to get the value of \[\tan \dfrac{\pi }{8}\].
Substituting \[A = \dfrac{\pi }{8}\] in the formula \[\cos 2A = 2{\cos ^2}A - 1\], we can find that \[{\cos ^2}\dfrac{\pi }{8} = \dfrac{{2 + \sqrt 2 }}{4}\].
Substituting \[{\cos ^2}\dfrac{\pi }{8} = \dfrac{{2 + \sqrt 2 }}{4}\] in the formula \[{\sin ^2}x + {\cos ^2}x = 1\], we can find that \[{\sin ^2}\dfrac{\pi }{8} = \dfrac{{2 - \sqrt 2 }}{4}\].
Dividing \[{\sin ^2}\dfrac{\pi }{8} = \dfrac{{2 - \sqrt 2 }}{4}\] by \[{\cos ^2}\dfrac{\pi }{8} = \dfrac{{2 + \sqrt 2 }}{4}\], we get
\[\begin{array}{l} \Rightarrow \dfrac{{{{\sin }^2}\dfrac{\pi }{8}}}{{{{\cos }^2}\dfrac{\pi }{8}}} = \dfrac{{\dfrac{{2 - \sqrt 2 }}{4}}}{{\dfrac{{2 + \sqrt 2 }}{4}}}\\ \Rightarrow {\tan ^2}\dfrac{\pi }{8} = \dfrac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }}\end{array}\]
Rationalising the denominator, we get
\[\begin{array}{l} \Rightarrow {\tan ^2}\dfrac{\pi }{8} = \dfrac{{2 - \sqrt 2 }}{{2 + \sqrt 2 }} \times \dfrac{{2 - \sqrt 2 }}{{2 - \sqrt 2 }}\\ \Rightarrow {\tan ^2}\dfrac{\pi }{8} = \dfrac{{4 + 2 - 4\sqrt 2 }}{{4 - 2}}\\ \Rightarrow {\tan ^2}\dfrac{\pi }{8} = \dfrac{{6 - 4\sqrt 2 }}{2}\\ \Rightarrow {\tan ^2}\dfrac{\pi }{8} = 3 - 2\sqrt 2 \end{array}\]
Taking square root of both sides, we get
\[ \Rightarrow \tan \dfrac{\pi }{8} = \sqrt {3 - 2\sqrt 2 } \]
This value of \[\tan \dfrac{\pi }{8}\] looks different from \[ - 1 + \sqrt 2 \], but they are equal because the actual values of both \[ - 1 + \sqrt 2 \] and \[\sqrt {3 - 2\sqrt 2 } \] is approximately \[{\rm{0}}{\rm{.41421356}}\].