# Greatest value of $\frac{1}{4\sin \theta-3 \cos \theta+6}$

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

Can anyone give me some hints on this?

• You can try to use derivation in order to find the extremal values of this function. – Alberto Debernardi Jan 12 '16 at 14:30
• Hint : $3^2 + 4^2 = 5^2$ – Jasser Jan 12 '16 at 14:30
• Is it $1$ the minimum value – Archis Welankar Jan 12 '16 at 14:34

Hint: Use $$-\sqrt {a^2+b^2} \le a\sin \theta + b \cos \theta \le \sqrt {a^2+b^2}$$.

Hints:

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

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

• Can you explain in detail? Thanks – Mathxx Jan 12 '16 at 14:35
• 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. – Deepak Jan 12 '16 at 14:38

$$4\sin\theta-3\cos\theta$$ 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 $$\frac1{-5+6}.$$