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If $\cos x=-\large\frac{\sqrt{15}}{4}$ and $\large\frac{\pi}{2}$$ \lt x $$ \lt \pi$ ,find the value of $\sin x$

$\begin{array}{1 1}(A)\;\large\frac{1}{2}&(B)\;\large\frac{1}{4}\\(C)\;\large\frac{1}{3}&(D)\;\large\frac{1}{5}\end{array} $

1 Answer

  • $\sin ^2x+\cos ^2x=1$
  • $\sin x=\sqrt{1-\cos^2x}$
Clearly $x$ lies in the II quadrant in which $\sin x$ and $cosec x$ are positive and all the other trigonometric functions are negative.
Now,$\cos x=-\large\frac{\sqrt{15}}{4}$
$\sin x=\sqrt{1-\cos ^2x}$
$\Rightarrow \sqrt{1-(-\large\frac{\sqrt 5}{4})^2}$
$\Rightarrow \sqrt{1-\large\frac{1 5}{16}}$
$\Rightarrow \sqrt{\large\frac{16-15}{16}}$
$\Rightarrow \large\frac{1}{4}$
$\therefore \sin x=\large\frac{1}{4}$
Hence (B) is the correct answer.
answered Apr 22, 2014 by sreemathi.v

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