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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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If $\theta$ lies in the first quadrant and $\cos\theta=\large\frac{8}{17}$,then the value of $\cos(30^{\large\circ}+\theta)+\cos(45^{\large\circ}-\theta)+\cos(120^{\large\circ}-\theta)$ ?

$\begin{array}{1 1}(A)\;\large\frac{23(\sqrt 3-\sqrt 2-1)}{34}&(B)\;\large\frac{23(\sqrt 3+\sqrt 2+1)}{34}\\(C)\;\large\frac{34(\sqrt 3-\sqrt 2-1)}{23}&(D)\;\large\frac{23(\sqrt 2-\sqrt 3-4)}{34}\end{array} $

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1 Answer

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Toolbox:
  • $\cos(a\pm b)=\cos a\cos b\mp \sin a.\sin b$
  • $\cos(\large\frac{\pi}{2}$$+x)=-\sin x$
$\theta$ lies in the first quadrant
$\cos\theta=\large\frac{8}{17}$
$\sin\theta=\large\frac{15}{17}$
$\tan\theta=\large\frac{15}{8}$
We have
$\cos(30^{\large\circ}+\theta)+\cos(45^{\large\circ}-\theta)+\cos(120^{\large\circ}-\theta)$
$\cos 30^{\large\circ}.\cos\theta-\sin 30^{\large\circ}.\sin \theta+\cos 45^{\large\circ}.\cos\theta+\sin 45^{\large\circ}.\sin \theta+\cos 120^{\large\circ}.\cos\theta+\sin 120^{\large\circ}\sin \theta$
$\large\frac{\sqrt 3}{2}\times \frac{8}{17}-\frac{1}{2}\times \frac{15}{17}+\frac{\sqrt 2}{2}.\frac{8}{17}+\frac{\sqrt 2}{2}.\frac{15}{17}+\frac{-1}{2}\times \frac{8}{17}+\frac{\sqrt 3}{2}\times \frac{15}{17}$
$\large\frac{4\sqrt 3}{17}-\frac{15}{34}+\frac{4\sqrt 2}{17}+\frac{15\sqrt 2}{34}-\frac{4}{17}+\frac{15\sqrt 3}{34}$
$\large\frac{8\sqrt 3-15+8\sqrt 2+15\sqrt 2-8+15\sqrt 3}{34}$
$\large\frac{23\sqrt 3-23\sqrt 2-23}{34}$
$\large\frac{23(\sqrt 3-\sqrt 2-1)}{34}$
Hence (A) is the correct answer.
answered Apr 20, 2014 by sreemathi.v
 

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