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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Evaluate $\sin 105^{\large\circ}+\cos 105^{\large\circ}$

$\begin{array}{1 1}(A)\;1&(B)\;\large\frac{1}{\sqrt 2}\\(C)\;\large\frac{1}{\sqrt 3}&(D)\;0\end{array} $

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1 Answer

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  • $\sin (A+B)=\sin A\cos B+\cos A\sin B$
We have
$\sin 105^{\large\circ}+\cos 105^{\large\circ}$
$\Rightarrow \sin (60^{\large\circ}+45^{\large\circ}+\cos(60^{\large\circ}+45^{\large\circ})$
$\Rightarrow (\sin 60^{\large\circ}\cos 45^{\large\circ}+\cos 60^{\large\circ}\sin 45^{\large\circ})+(\cos 60^{\large\circ}\cos 45^{\large\circ}-\sin 60^{\large\circ}\sin 45^{\large\circ})$
$\Rightarrow [(\large\frac{\sqrt 3}{2}\times \frac{1}{\sqrt 2})+(\large\frac{1}{2}\times \frac{1}{\sqrt 2})]+[(\large\frac{1}{2}\times \frac{1}{\sqrt 2})-(\large\frac{\sqrt 3}{2}\times \frac{1}{\sqrt 2})]$
$\Rightarrow \big[\large\frac{\sqrt 3}{2\sqrt 2}+\frac{1}{2\sqrt 2}+\large\frac{1}{2\sqrt 2}- \frac{\sqrt 3}{2\sqrt 2}\big]$
$\Rightarrow \large\frac{1}{\sqrt 2}$
Hence (B) is the correct answer.
answered May 21, 2014 by sreemathi.v
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