Find the value of
Question:

Find the value of $\sin 405^{\circ}$

Solution:

To find: Value of sin 405°

We have,

$\sin 405^{\circ}=\sin \left[90^{\circ} \times 4+45^{\circ}\right]$

$=\sin 45^{\circ}$

[Clearly, $405^{\circ}$ is in ${ }^{\text {st }}$ Quadrant and the multiple of $90^{\circ}$ is even]

$=\frac{1}{\sqrt{2}}\left[\because \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right]$