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The question says to find the range and domain of $$P(x) = \frac{\sin(x)-1}{\sqrt{3-2\sin(x)-2\cos(x)}}$$

How do I approach this problem? For domain, I know I should set the denominator $>0$ so that it doesn't become undefined...but not really how to proceed further.

Any help would be appreciated.

N. F. Taussig
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Techie5879
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1 Answers1

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Notice that

$$3-2\sin(x)-2\cos(x) = 2 + \sin^2(x)+\cos^2(x)-2\sin(x)-2\cos(x)$$ $$= (\sin^2(x) - 2\sin(x) + 1) + (\cos^2(x) - 2\cos(x) + 1) = (\sin(x)-1)^2 + (\cos(x)-1)^2$$

In other words our function can be rewritten as

$$P(x) = \frac{\sin(x)-1}{\sqrt{(\sin(x)-1)^2 + (\cos(x)-1)^2}}$$

The numerator is always nonpositive, at most $0$. To get the minimum negative value,

$$\frac{\sin(x)-1}{\sqrt{(\sin(x)-1)^2 + (\cos(x)-1)^2}} \geq \frac{\sin(x)-1}{\sqrt{(\sin(x)-1)^2}} = -1$$

which is okay because both squares inside the square root cannot be $0$ at the same time.

So the range of this function is $[-1,0]$.

Ninad Munshi
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  • Could you please explain the last step....how you got the minimum negative value – Techie5879 Sep 06 '19 at 08:05
  • @Techie5879 Take absolute value instead and denote $a = 1-\sin(x)$ and $b=1-\cos(x)$. Then $\frac{a}{\sqrt{a^2+b^2}} \leq \frac{a}{\sqrt{a^2}} = 1$. The minus sign flips the inequality. – Ninad Munshi Sep 06 '19 at 08:10