# Range of the trigonometric function $f(x)= \frac{3}{4 - 5\sin(x)}$

I was trying to find the range of the function

$$f(x) = \frac{3}{4-5\sin(x)}$$

The first thought that came to mind was:

Let's put the minimum and maximum value of $$\sin(x)$$ as $$\{1,-1\}$$ accordingly to find maximum and minimum of the function.

However, then I got range $$[-3,\frac{1}{3}]$$, which was totally wrong when I used a graphing calculator.

The true answer is $$(-\infty,-3] ∪ [\frac{1}{3}, \infty)$$

Now, I want explaination what I did wrong here.

• If $f(x)$ were defined and continuous everywhere, your approach would work. But look at what happens whenever $\sin x=4/5$. Feb 6 at 7:54

Let the denominator be $$y$$. Then, as you observed $$-1\leq y\leq 9$$. But you have to be careful when find the range for $$\frac 1 y$$. At $$y=0$$, $$\frac 1 y$$ is not defined. You have to consider $$0 and $$-1\leq y<0$$ separately. For $$0 we get $$\frac 1 y \in [\frac 1 9, \infty)$$ and for $$-1\leq y<0$$ we get $$\frac 1 y \in (-\infty, -1]$$. Now multiply by $$3$$.

To really understand how to avoid arriving at an incorrect conclusion, it is important to consider not just the extrema of the interval that $$\sin x$$ maps to, but also the intermediate values that are possible. To do this, observe $$-1 \le \sin x \le 1 \tag{1}$$ implies $$5 \ge -5 \sin x \ge -5. \tag{2}$$ (Recall that multiplying an inequality by a negative number reverses the direction of the inequality.) Hence $$4 + 5 \ge 4 - 5 \sin x \ge 4 - 5$$ or equivalently $$9 \ge 4 - 5 \sin x \ge -1. \tag{3}$$ Now, here is where the mistake is made: if you took reciprocals of this inequality, you might think it will look like this:

$$\frac{1}{9} \le \frac{1}{4 - 5 \sin x} \le -1.$$ Again, remember that taking reciprocals, like negation, will reverse the direction of the inequality. But this inequality does not make any sense because $$1/9 > -1$$. In the previous step $$(3)$$, clearly the choice $$x = 0$$ would give us a valid result, $$9 \ge 4 \ge -1$$. But after taking reciprocals, $$1/9 \le 1/4$$ is still true, but $$1/4 \le -1$$ is false. A piece of reasoning is missing.

The missing piece is that the inequality $$(3)$$ contains $$0$$. If I had written $$9 \ge y \ge -1,$$ then $$y = 0$$ is a valid solution to this inequality; however, upon taking reciprocals, $$1/y$$ is not defined when $$y = 0$$. So this is the first hint. The second hint is that if $$y$$ is negative, say $$y = -1/2$$, then taking reciprocals gives us $$9 \le -2 \le -1$$ which is true on the right-hand side but not the left. So what we need to do here is split the compound inequality $$(3)$$ into separate inequalities according to whether the middle term is positive or negative (zero is not allowed):

$$9 \ge 4 - 5 \sin x > 0, \quad {\bf\text{or}} \quad 0 > 4 - 5 \sin x \ge -1. \tag{4}$$ Now that we have done this, taking reciprocals of each will no longer result in a contradiction:

$$\frac{1}{9} \le \frac{1}{4 - 5 \sin x} < \infty, \quad \text{or} \quad -\infty < \frac{1}{4 - 5 \sin x} \le -1. \tag{5}$$ Now we multiply by $$3$$ and drop the infinities (because they are implied) to get $$\frac{1}{3} \le \frac{3}{4 - 5 \sin x}, \quad \text{or} \quad \frac{3}{4 - 5 \sin x} \le -3. \tag{6}$$ This gives us the desired range of the function.

Although somewhat convoluted, a nice approach to finding range could be by finding the domain of the inverse function.

Notice that if $$f(x)={{3}\over{4-5\sin{x}}}$$ then $$f^{-1}(x)=\sin^{-1}\big({4\over5}-{3\over{5x}}\big)$$ making the domain of $$f^{-1}(x)$$ satisfying: $$-1\leq{4\over5}-{3\over{5x}}\leq1$$ which simplifies to: $$3\geq{1\over x}\geq-{1\over3}$$ Looking at where equality holds, we get $$x={1\over3}$$ or $$x=-3$$. Testing for when the inequality above holds by using test values around $$x={1\over3}$$ and $$x=-3$$ on the entire real line (say $$x=-4,\ -1,\ 1$$) we get that $$x\in(-\infty,-3]\cup\big[{1\over3},\infty\big)$$, which is also the range of $$f(x)$$, as desired.

Instead, look at the range of the function $$g(x)=\frac{1}{f(x)}=\frac{4-5\sin(x)}{3}$$ What is the range of $$g(x)$$?

$$-1\le\sin(x)\le1\Rightarrow\frac{-1}{3}\le\frac{4-5\sin(x)}{3}\le3$$ Hence, range of $$g(x)$$ is $$\left[\frac{-1}{3},3\right]$$. What should be the range of the inverse of this?

$$(-\infty,-3]\cup[\frac{1}{3},\infty)$$ Note that for $$f(x)$$ to exist, $$\sin(x)\ne\frac{4}{5}$$. Make sure the domain doesn't include this particular $$x$$.