Question

How do you evaluate $\sin$, $\cos$, tangent of $\frac{\pi }{4}$ without using a calculator?

Answer 1

Step-1: Finding the value of $\sin \left(\frac{\pi }{4}\right)$:

$\cos \left(2\frac{\pi }{4}\right)=1-2{\sin }^2\left(\frac{\pi }{4}\right)$ $\left[\because \cos \left(2\theta \right)=1-2{\sin }^2\left(\theta \right)\right]$

$\Rightarrow$ $\cos \left(\frac{\pi }{2}\right)=1-2{\sin }^2\left(\frac{\pi }{4}\right)$

$\Rightarrow$ $0=1-2{\sin }^2\left(\frac{\pi }{4}\right)$ $\left[\because \cos \left(\frac{\pi }{2}\right)=0\right]$

$\Rightarrow$ ${\sin }^2\left(\frac{\pi }{4}\right)=\frac{1}{2}$

$\Rightarrow$ $\sin \left(\frac{\pi }{4}\right)=\pm \sqrt{\frac{1}{2}}$

$\Rightarrow$ $\sin \left(\frac{\pi }{4}\right)=+\sqrt{\frac{1}{2}}$ [$\because \frac{\mathrm{\pi }}{4}$ is in $1^{st}$ quadrant]

$\therefore \sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$

Step-2: Finding the value of $\cos \left(\frac{\pi }{4}\right)$:

$\cos \left(\frac{\pi }{4}\right)=\pm \sqrt{1-{\sin }^2\left(\frac{\pi }{4}\right)}$ $\left[\because \cos \left(\theta \right)=\sqrt{1-{\sin }^2\left(\theta \right)}\right]$

$=+\sqrt{1-{\sin }^2\left(\frac{\mathrm{\pi }}{4}\right)}$ [$\because \frac{\mathrm{\pi }}{4}$ is in $1^{st}$ quadrant]

$=\sqrt{1-{\left(\frac{1}{\sqrt{2}}\right)}^2}$

$=\sqrt{1-\frac{1}{2}}$

$=\frac{1}{\sqrt{2}}$

$\therefore \sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$

Step-3: Finding the value of $\tan \left(\frac{\pi }{4}\right)$:

$\tan \left(\frac{\pi }{4}\right)=\frac{\sin \left({\displaystyle \frac{\pi }{4}}\right)}{\cos \left({\displaystyle \frac{\pi }{4}}\right)}$ $\left[\because \tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\right]$

$=\frac{{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}}$ $\left[\because \sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}},\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\right]$ $=1$

$\therefore \tan \left(\frac{\mathrm{\pi }}{4}\right)=1$

Hence, the value of $\sin \left(\frac{\pi }{4}\right)$ is $\frac{1}{\sqrt{2}}$, $\cos \left(\frac{\pi }{4}\right)$ is $\frac{1}{\sqrt{2}}$ and $\tan \left(\frac{\pi }{4}\right)$ is $1$.