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Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $

Find the minimum value of $h(\theta)$

$h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} = 5 \sin (\theta + 53.13) + \sqrt{2} $

Minimum value -

$5\sin (\theta + 53.13) + \sqrt{2} = -5 $

Therefore min value is = $ -5/5 - \sqrt{2} $

Why am I wrong ? And how should I do this question..

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  • $\begingroup$ I would say that the standard way is finding the derivative of the function and solving for $h'(\theta) = 0$. But algebraic manipulation can work too. Numerically, I find the minimum to be $\approx -3.5858$. $\endgroup$ – Matti P. Feb 19 '18 at 7:13
  • $\begingroup$ @user175089...Find h'(θ) and h''(θ)...Take h'(θ)=0...For the value of θ for which h''(θ)>0 ,gives the minimum value... $\endgroup$ – Nehal Samee Feb 19 '18 at 7:14
  • $\begingroup$ You made a sign error. $-5\color{red}+\sqrt2$. And why do you divide by $5$ ?? $\endgroup$ – Yves Daoust Feb 19 '18 at 7:14
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Minimum is attained when $\sin (\theta + 53.13) = -1$ that is

$$h_{min}=h(3\pi/2+k\pi)=5 \sin (3\pi/2+k\pi) + \sqrt{2}=-5+\sqrt{2}$$

for the same reason the maximum is attained when $\sin (\theta + 53.13) = 1$.

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Hint

You must note that the range of $a\sin\theta \pm b\cos\theta$ is $\left[ -\sqrt {a^2+b^2}, \sqrt {a^2+b^2}\right]$

Hence in your case range of given function is $[-5+\sqrt 2, 5+\sqrt 2]$

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Apply Buniakowski inequality: $(3\sin \theta + (-4)\cos \theta)^2 \le (3^2+(-4)^2)(\sin^2 \theta+\cos^2 \theta) = 25\implies 3\sin \theta - 4\cos \theta \ge -5\implies $ min value $ = -5+\sqrt{2}$

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    $\begingroup$ I don't know about Buniakowski inequality but It is simply Cauchy Schwarz inequality $\endgroup$ – Rohan Shinde Feb 19 '18 at 7:21

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