Prove that the value of $5\cos \theta+3\cos (\theta+\large\frac{\pi}{3})+$$3 lies between -4 and 10 1 Answer Toolbox: • a\cos\theta+b\sin \theta are in the form k\cos \phi or k\sin \phi • k=\sqrt{a^2+b^2} • \cos(A+B)=\cos A\cos B+\sin A\sin B Let Y=5\cos \theta+3\cos(\theta+\large\frac{\pi}{3})$$+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