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

Question

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..

Answer 1

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 $\sin (\theta + 53.13) = 1$.

Answer 2

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 $[-5+\sqrt 2, 5+\sqrt 2]$

Answer 3

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}$