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