# 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..

• Well, you need to identify $a$ and $b$ (done!), square them (done!), add the squares (done!), and then take the square root of the sum (?). – DJohnM Jul 4 '13 at 5:29
• I have the answer which is $-2(2+\sqrt{5}) , 2(2+\sqrt{5})$ I don't understand how it came...please guide.. – Sachin Jul 4 '13 at 5:47

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 $[-\sqrt{a^2+b^2},\, \sqrt{a^2+b^2}]$

Note that $-1 \leq \sin \theta, \cos \theta \leq 1$ so $-2\sqrt 3 -4-4\leq f(\theta) \leq 2\sqrt 3 +4 +4$ You still need to worry about the correlations between $\sin$ and $\cos$.