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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove that the value of $5\cos \theta+3\cos (\theta+\large\frac{\pi}{3})+$$3$ lies between -4 and 10

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Toolbox:
  • $a\cos\theta+b\sin \theta$ are in the form $k\cos \phi$ or $k\sin \phi$
  • $k=\sqrt{a^2+b^2}$
  • $\cos(A+B)=\cos A\cos B+\sin A\sin B$
Let $Y=5\cos \theta+3\cos(\theta+\large\frac{\pi}{3})$$+3$
$\Rightarrow 5\cos \theta+3(\cos \theta\cos\large\frac{\pi}{3}$$-\sin \theta\sin \large\frac{\pi}{3})$$+3$
$\Rightarrow 5\cos\theta+\large\frac{3}{2}$$\cos \theta-\large\frac{3\sqrt 3}{2}$$\sin \theta+3$
$\Rightarrow \large\frac{13}{2}$$\cos \theta-\frac{3\sqrt 3}{2}$$\sin \theta+2$
$\Rightarrow 7(\large\frac{13}{14}$$\cos \theta-\large\frac{3\sqrt 3}{14}$$\sin \theta)+3$
Dividing & multiplying by
$(\sqrt{\large\frac{13}{2})^2+(\frac{-3\sqrt 3}{2})^2}=$$7$
Let $\cos \alpha=\large\frac{13}{14}$,then $\sin \alpha=\large\frac{3\sqrt 3}{14}$
Now $y=7(\cos \alpha\cos \theta-\sin \alpha\sin \theta)+3$
$\Rightarrow 7\cos (\theta+\alpha)+3$
Maximum value of $\cos(\theta+\alpha)=1$
Maximum value of y $7.1+3=10$
Minimum value of $\cos(\theta+\alpha)=-1$
Minimum value of y $7.(-1)+3=-4$
$5\cos \theta+3\cos (\theta+\large\frac{\pi}{3})+$$3$ lies between -4 and 10
Hence proved
answered May 21, 2014 by sreemathi.v
 

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