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.. 

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