<s>[INST] 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.. 
 [/INST] 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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