<s>[INST] Trigonometry - Finding the range of the function Problem : 
$$f(\theta)=(2\sqrt{3}+4)\sin\theta +4\cos \theta $$
I have studied if the function is in the form : $f(\theta)=a\cos\theta + b\sin\theta$ then the range of this function can be given as $$[-\sqrt{a^2+b^2} ,\sqrt{a^2+b^2}]$$
So, here  the range will be [-$44+16\sqrt{3} , 44+16\sqrt{3}]$ But this is the wrong answer.. please guide.. thanks..
 [/INST] The answer is indeed  $ [-\sqrt{a^2+b^2},\, \sqrt{a^2+b^2}]$. 
You made a calculation mistake somewhere. The range is $[-\sqrt{44+16\sqrt{3}},\sqrt{44+16\sqrt{3}}]$
Here's a proof:
Let $a=2\sqrt{3}+4$ and $b= 4$
Choose $y \in [0,2\pi]$ such that  $\sin y = \dfrac{b}{\sqrt{a^2+b^2}}$ and $\cos y = \dfrac{a}{\sqrt{a^2+b^2}}$
Then $f(\theta) = a\sin\theta + b\cos\theta = \sqrt{a^2+b^2}(\dfrac{a}{\sqrt{a^2+b^2}}\sin\theta + \dfrac{b}{\sqrt{a^2+b^2}}\cos\theta) = \sqrt{a^2+b^2}(\cos y\,\sin\theta+\sin y\,\cos \theta) = \sqrt{a^2+b^2}\,\sin(\theta+y)$
which lies in the range $ [-\sqrt{a^2+b^2},\, \sqrt{a^2+b^2}]$
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