Question

Solve the below trigonometric question.

Find the value of tan(π/8).

This is trigonometric functions chapter.

Answer 1

$\large\underline{\sf{Solution-}}$

Given trigonometric function is

$\rm \: tan\bigg(\dfrac{\pi}{8} \bigg)$

We know,

$\rm \: tan2x = \dfrac{2tanx}{1 - {tan}^{2}x}$

$\red{ \sf \: Put \: x \: = \: \dfrac{\pi}{8} \: in \: the \: above \: expression, \: we \: get}$

$\rm \: tan2\bigg(\dfrac{\pi}{8} \bigg) = \dfrac{2tan\bigg(\dfrac{\pi}{8} \bigg)}{1 - {tan}^{2}\bigg(\dfrac{\pi}{8} \bigg)}$

$\rm \: tan\bigg(\dfrac{\pi}{4} \bigg) = \dfrac{2tan\bigg(\dfrac{\pi}{8} \bigg)}{1 - {tan}^{2}\bigg(\dfrac{\pi}{8} \bigg)}$

$\rm \: 1 = \dfrac{2tan\bigg(\dfrac{\pi}{8} \bigg)}{1 - {tan}^{2}\bigg(\dfrac{\pi}{8} \bigg)}$

$\rm \: 1 - {tan}^{2}\bigg(\dfrac{\pi}{8} \bigg) = 2tan\bigg(\dfrac{\pi}{8} \bigg)$

can be rewritten as

$\rm \: {tan}^{2}\bigg(\dfrac{\pi}{8} \bigg) + 2tan\bigg(\dfrac{\pi}{8} \bigg) - 1 = 0$

Its a quadratic equation in $\rm \: tan\bigg(\dfrac{\pi}{8} \bigg)$, so using quadratic formula we have,

$\rm \: tan\bigg(\dfrac{\pi}{8} \bigg) = \dfrac{ - 2 \: \pm \: \sqrt{ {2}^{2} - 4(1)( - 1)} }{2(1)}$

$\rm \: tan\bigg(\dfrac{\pi}{8} \bigg) = \dfrac{ - 2 \: \pm \: \sqrt{4 + 4} }{2}$

$\rm \: tan\bigg(\dfrac{\pi}{8} \bigg) = \dfrac{ - 2 \: \pm \: \sqrt{8} }{2}$

$\rm \: tan\bigg(\dfrac{\pi}{8} \bigg) = \dfrac{ - 2 \: \pm \: 2 \sqrt{2} }{2}$

$\rm \: tan\bigg(\dfrac{\pi}{8} \bigg) = - 1\: \pm \: \sqrt{2}$

As,

$\rm \: 0 < \bigg(\dfrac{\pi}{8} \bigg) < \bigg(\dfrac{\pi}{2} \bigg)$

$\bf\implies \:tan\bigg(\dfrac{\pi}{8} \bigg) > 0$

$\bf\implies \:tan\bigg(\dfrac{\pi}{8} \bigg) = \sqrt{2} - 1$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered} { \boxed{ \begin{array}{c} \underline{\underline{ \color{orange} \text{Additional \: lnformation}}} \\& \rm \: sin2x \: = 2 \: sinx \: cosx\:\\ & \rm \: cos2x = 1 - {2sin}^{2}x \\ & \rm \: cos2x = {2cos}^{2}x - 1 \\ & \rm \: cos2x = {cos}^{2}x - {sin}^{2}x \\ & \rm \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} \\ & \rm \: sin2x = \dfrac{2tanx}{1 + {tan}^{2}x } \\ & \rm \:sin3x = 3sinx - {4sin}^{3}x \\ & \rm \: cos3x = {4cos}^{3}x - 3cosx \\ & \rm \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - {3tan}^{2}x} \end{array}}}\end{gathered}$

Answer 2

Step-by-step explanation:

$to \: find : \\ tan( \frac{\pi}{8} ) = tan(22.5) \\ \\ we \: know \: that \\ \frac{45}{2} = 22.5 \\ \\ taking \: tangent \: function \: on \\ \: both \: sides \\ \\ \tan( \frac{45}{2} ) = \tan(22.5) \\ \\ we \: know \: that \\ tan( \frac{x}{2} ) = \sqrt{ \frac{1 - cos \: x}{1 + cos \: x} } = cosec(x) - cot(x) \\ \\ = cosec(45) - cot(45) \\ \\tan(22.5) = tan( \frac{\pi}{ 8 }) = \sqrt{2} - 1$