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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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The value of $\sin\large\frac{\pi}{18}$$+\sin \large\frac{\pi}{9}$$+\sin \large\frac{2\pi}{9}$$+\sin \large\frac{5\pi}{18}$ is given by

$\begin{array}{1 1}(A)\;\sin\large\frac{7\pi}{18}+\normalsize \sin \large\frac{4\pi}{9}&(B)\;1\\(C)\;\cos \large\frac{\pi}{6}+\normalsize \cos \large\frac{3\pi}{7}&(D)\;\cos \large\frac{\pi}{9}+\normalsize \sin \large\frac{\pi}{9}\end{array} $

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1 Answer

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  • $\sin A+\sin B=2\sin \large\frac{A+B}{2}$$.\cos \large\frac{A-B}{2}$
$\sin\large\frac{\pi}{18}$$+\sin \large\frac{\pi}{9}$$+\sin \large\frac{2\pi}{9}$$+\sin \large\frac{5\pi}{18}$
$\Rightarrow \sin\large\frac{5\pi}{18}$$+\sin \large\frac{\pi}{18}$$+\sin \large\frac{2\pi}{9}$$+\sin \large\frac{\pi}{9}$
$\sin A+\sin B=2\sin \large\frac{A+B}{2}$$.\cos \large\frac{A-B}{2}$
$\Rightarrow 2\sin\large\frac{\pi}{6}.$$\cos\large\frac{\pi}{9}$$+2\sin \large\frac{\pi}{6}$$\cos\large\frac{\pi}{18}$
$\cos \large\frac{\pi}{9}$$+\cos \large\frac{\pi}{18}$
$\cos(\large\frac{\pi}{2}-\frac{7\pi}{8})+$$\cos (\large\frac{\pi}{2}-\frac{4\pi}{9})$
$\sin \large\frac{7\pi}{18}$$+\sin \large\frac{4\pi}{9}$
Hence (A) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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