Question

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

Answer 1

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 ${\rm{0}}{\rm{.41421356}}$.