5
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State the greatest value of $$\frac{1}{4\sin \theta-3 \cos \theta+6}$$

Can anyone give me some hints on this?

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  • 1
    $\begingroup$ You can try to use derivation in order to find the extremal values of this function. $\endgroup$ – Alberto Debernardi Jan 12 '16 at 14:30
  • $\begingroup$ Hint : $3^2 + 4^2 = 5^2$ $\endgroup$ – Jasser Jan 12 '16 at 14:30
  • $\begingroup$ Is it $1$ the minimum value $\endgroup$ – Archis Welankar Jan 12 '16 at 14:34
11
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Hint: Use $$-\sqrt {a^2+b^2} \le a\sin \theta + b \cos \theta \le \sqrt {a^2+b^2}$$.

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5
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Hints:

$4\sin\theta - 3\cos\theta = 5\sin(\theta - \arctan\frac 34)$

$-1 \leq \sin\alpha \leq 1$

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  • $\begingroup$ Can you explain in detail? Thanks $\endgroup$ – Mathxx Jan 12 '16 at 14:35
  • $\begingroup$ Do you know the manipulation in the first step? Any expression of the form $a\sin \theta \pm b\cos\theta$ can be expressed as $\sqrt{a^2 + b^2} \sin (\theta \pm \arctan \frac ba)$. The second part uses the bounds of the sine function. To maximise the ratio, find the minimum possible value of the denominator. Now figure out what that must be. $\endgroup$ – Deepak Jan 12 '16 at 14:38
3
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$$4\sin\theta-3\cos\theta$$ is the dot product of the vector $(-3,4)$ with the unit vector in direction $\theta$. This dot product is minimized when the two vectors are antiparallel, and equals minus the product of the norms, i.e. $-5$.

The requested maximum is $$\frac1{-5+6}.$$

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