Question

Find the value of

$cot \frac{\pi}{8}$

​

Answer 1

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

Given expression is

$\rm \: cot \: \dfrac{\pi}{8}$

Let assume that,

$\rm \: \: \dfrac{\pi}{8} = x$

So, above can be rewritten as

$\rm \: cotx \: = \: \dfrac{cosx}{sinx}$

On multiply numerator and denominator by 2 cosx, we get

$\rm \: cotx \: = \: \dfrac{2cos^{2} x}{2 \: cosx \: sinx}$

We know,

$\boxed{\sf{ \:cos2x = {2cos}^{2}x - 1 \: \: }}$

and

$\boxed{\sf{ \:sin2x \: = \: 2 \: sinx \: cosx \: }}$

So, using these results, we get

$\rm \: cotx \: = \: \dfrac{1 + cos2x}{sin2x}$

On substituting the value of x, we get

$\rm \: cot\dfrac{\pi}{8} \: = \: \dfrac{1 + cos2\bigg(\dfrac{\pi}{8}\bigg)}{sin2\bigg(\dfrac{\pi}{8}\bigg)}$

$\rm \: cot\dfrac{\pi}{8} \: = \: \dfrac{1 + cos\dfrac{\pi}{4}}{sin\dfrac{\pi}{4}}$

$\rm \: cot\dfrac{\pi}{8} \: = \: \dfrac{1 + \dfrac{1}{ \sqrt{2} }}{\dfrac{1}{ \sqrt{2} }}$

$\rm \: cot\dfrac{\pi}{8} \: = \: \dfrac{\dfrac{ \sqrt{2} + 1}{ \sqrt{2} }}{\dfrac{1}{ \sqrt{2} }}$

Thus,

$\rm\implies \:\boxed{\sf{ \:\rm \: cot\dfrac{\pi}{8} \: = \: \sqrt{2} \: + \: 1 \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information

$\boxed{\sf{ \:\rm \: sin18\degree = \frac{ \sqrt{5} - 1}{4} \: }}$

$\boxed{\sf{ \:\rm \: cos18\degree = \frac{ \sqrt{10 + 2 \sqrt{5} } }{4} \: }}$

$\boxed{\sf{ \:\rm \: sin36\degree = \frac{ \sqrt{10 - 2 \sqrt{5} } }{4} \: }}$

$\boxed{\sf{ \:\rm \: cos54\degree = \frac{ \sqrt{10 - 2 \sqrt{5} } }{4} \: }}$

$\boxed{\sf{ \:\rm \: sin72\degree = \frac{ \sqrt{10 + 2 \sqrt{5} } }{4} \: }}$

$\boxed{\sf{ \:\rm \: cos72\degree = \frac{ \sqrt{5} - 1}{4} \: }}$

$\boxed{\sf{ \:\rm \: tan\dfrac{\pi}{8} \: = \: \sqrt{2} - 1 \: \: }}$

Answer 2

$\large\mathbb \blue{\fcolorbox{red}{yellow}{ missincredeible \: here\ }}$