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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the maximum and minimum values of the following expression $3\cos\theta+5\sin (\theta-\large\frac{\pi}{6})$

$\begin{array}{1 1}(A)\;Y_{max}=\sqrt {19},Y_{min}=-\sqrt{19}\\(B)\;Y_{max}=\sqrt {17},Y_{min}=-\sqrt{17}\\(C)\;Y_{max}=\sqrt {13},Y_{min}=-\sqrt{13}\\(D)\;Y_{max}=\sqrt {21},Y_{min}=-\sqrt{21}\end{array} $

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1 Answer

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Toolbox:
  • $a\cos \theta+b\sin \theta$ are in the form $k\cos\phi$ or $k\sin \phi$
  • $k=\sqrt{a^2+b^2}$
$y=3\cos \theta+5\sin (\theta-\large\frac{\pi}{6})$
$y=3\cos \theta+5(\sin \theta\cos\large\frac{\pi}{6}$$-\cos\theta\sin \large\frac{\pi}{6})$
$\Rightarrow 3\cos\theta+5(\large\frac{\sqrt 3}{2}$$\sin \theta-\large\frac{1}{2}$$\cos\theta)$
$\Rightarrow \large\frac{1}{2}$$(5\sqrt 3\sin \theta-\cos \theta)$
Multiplying and dividing by
$\sqrt{5\sqrt 3)^2+1^2}=\sqrt{76}$
$\Rightarrow \large\frac{1}{2}$$\sqrt{76}(\large\frac{5\sqrt 3}{\sqrt{76}}$$\sin \theta-\large\frac{1}{\sqrt{76}}$$\cos\theta)$
$\Rightarrow \sqrt{19}(\cos\alpha.\sin \theta-\sin \alpha\cos\theta)$
Where $\alpha=\tan-(\large\frac{1}{5\sqrt 3})$
Now maximum value of $\sin(\theta-\alpha)=1$
$\therefore Y_{max}=\sqrt{19}$
Now minimum value of $\sin(\theta-\alpha)=-1$
$\therefore Y_{min}=-\sqrt{19}$
Hence (A) is the correct answer.
answered May 21, 2014 by sreemathi.v
 

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