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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the value of $\sin 405^{\large\circ}$

$\begin{array}{1 1}(A)\;\large\frac{1}{\sqrt 2}&(B)\;\large\frac{1}{\sqrt 3}\\(C)\;\large\frac{1}{\sqrt 6}&(D)\;\sqrt 2\end{array} $

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1 Answer

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  • $\sin(\theta+2n\pi)=\sin \theta$
$180=\pi^c$
$405^{\large\circ}=(\large\frac{\pi}{180}$$\times 405)=\large\frac{9\pi}{4}$
$\sin 405^{\large\circ}=\sin (\large\frac{9\pi}{4})$
$\sin \large\frac{9\pi}{4}=$$\sin (2\pi+\large\frac{\pi}{4})$
$\sin \large\frac{\pi}{4}=\frac{\sqrt 2}{2}=\frac{1}{\sqrt 2}$
Hence (A) is the correct answer.
answered Apr 22, 2014 by sreemathi.v
 
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