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

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}$
