Question

Explain two methods to elucidate the value of $\cos \left(45°\right)$ using $\sin \left(x\right)$ and $\cos \left(x\right)$?

Answer 1

To find the value of $\cos \left({45}^{{\circ }^{}}\right)$.

Method: 1

Step 1: Use the relation ${\sin }^2\left(x^{}\right)+{\cos }^2\left(x\right)=1$

We know that,

${\cos }^2\left(x\right)+{\sin }^2\left(x\right)=1$

$\Rightarrow \cos \left(x\right)=\sqrt{1-{\sin }^2\left(x\right)}$

Step 2: Putting the value of $x$

we have,

$$\begin{gathered} \Rightarrow \cos {45}^{\circ }=\sqrt{1-{\sin }^2{\left(45\right)}^{\circ }} \\ \Rightarrow \cos {45}^{\circ }=\sqrt{1-(\frac{1}{\sqrt{2}}}{)}^2 \\ \Rightarrow \cos {45}^{\circ }=\frac{1}{\sqrt{2}} \end{gathered}$$

Method: 2

Step 1: Use the relation $\sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right)$

$\Rightarrow \cos \left(x\right)=\frac{\sin \left(2x\right)}{2\sin \left(x\right)}$

Step 2: Putting the value of $x$

$\begin{array}{cccc}\Rightarrow & \cos x& =& \frac{\sin \left(2x\right)}{2\sin \left(x\right)}\\ \Rightarrow & \cos 45°& =& \frac{\sin (2\times {45}^{\circ })}{2\sin \left({45}^{\circ }\right)}\\ \Rightarrow & \cos 45°& =& \frac{\sin \left({90}^{\circ }\right)}{2\sin \left({45}^{\circ }\right)}\\ \Rightarrow & \cos 45°& =& \frac{1}{2\times {\displaystyle \frac{1}{\sqrt{2}}}}\\ \Rightarrow & \cos 45°& =& \frac{1}{\sqrt{2}}\end{array}$

Hence, the value of $\cos \left(45°\right)$ is $\frac{1}{\sqrt{2}}$.