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Find the range of the function $$f(x) = \frac{2 + \sin x}{3 + \cos x}.$$

Honestly, I'm completely stumped. This is a pre-calculus course. We haven't learned derivatives and we are not allowed to use a calculator. I understand the range to be the distance between the minimum and maximum of the function. Theoretically I was supposing it might be like tangent with a range of $-\infty$ to $+\infty$. I know a sine curve's range is $[-1, 1]$ as is cosine, but I do not know what to do about combining them. My teacher is of very little use and gives us questions beyond what he's taught.

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You don't need calculus to find the range.

Write $$y=\frac{2+\sin x}{3+\cos x}$$ and rearrange as $$\sin x-y\cos x=3y-2$$

Use a compound angle transformation to rewrite the left hand side as $$\sqrt{1+y^2}\sin(x+\alpha)$$

Knowing the range of $\sin$ enables us to write $$\left|\frac{3y-2}{\sqrt{1+y^2}}\right|\leq 1$$

Squaring this and rearranging leads to $$8y^2-12y+3\leq0$$

So we end up with the range which is $$\frac 14(3-\sqrt{3})\leq y\leq \frac 14(3+\sqrt{3})$$

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  • $\begingroup$ This Desmos graph gives some graphical evidence the above calculations are correct: desmos.com/calculator/zfr0ydylxp $\endgroup$ Jul 22 '19 at 20:35
  • $\begingroup$ Can you explain "Use a compound angle transformation to rewrite the left hand side as" this step to me? How did you transform first equation? $\endgroup$ Jun 22 '20 at 15:59
  • $\begingroup$ @NikaChelidze Let $$\sin x-y\cos x=R\sin(x+\alpha)$$ and get $R$ in terms of $y$ $\endgroup$ Jun 23 '20 at 12:51

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