Arithmetic Progression: $ \log(2), \log(2\sin(x)-1),\log(1-y)$ , Solve for possible values of $y$

Question

Question: An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

The following numbers form three consecutive terms of an arithmetic sequence, for some real $x$ and $y$$:$

$$ \log(2), \log(2\sin(x)-1),\log(1-y)$$

Find the range of possible values of $y$.

What I have attempted:

I know that in an arithmetic progression such as

$$ a,b,c$$

that

$$ b-a = c-b $$

$$ 2b = a+c$$

$$ b = \frac{a+c}{2}$$

So by using this formula in the given sequence

$$ \frac{\log(2) + \log(1-y)}{2} = \log(2\sin(x)-1)$$

$$ \log(2(1-y) = 2\log(2\sin(x)-1) $$

$$ \log(2-2y) = \log((2\sin(x)-1)^2) $$

$$ 2-2y = (2\sin(x)-1)^2$$

$$ 2-(2\sin(x)-1)^2 = 2y $$

$$ y = \frac{2-(2\sin(x)-1)^2 }{2} $$

But here I am stuck , how am I suppose to find the range of values of $y$ , because there is a $\sin(x)$?

Answer 1

Severely edited:

$$-1 \le \sin(x) \le 1$$ $$-2 \le 2\sin(x) \le 2$$ $$-3 \le 2\sin(x)-1 \le 1$$

but we may be restricted to cases where $0 \lt 2\sin(x)-1$ so that we can take the real logarithm so

$$0 \lt 2\sin(x)-1 \le 1$$

and $\log(a),\log(b),\log(c)$ are in arithmetic progression when $a,b,c$ are in geometric progression, i.e. when $a = \frac{b^2}{c}$ and $c = \frac{b^2}{a}$, and this means

$$0 \lt (2\sin(x)-1)^2 \le 1$$ $$0 \lt \frac{(2\sin(x)-1)^2}{2} \le \frac12$$ $$\frac12 \le 1-\frac{(2\sin(x)-1)^2}{2} \lt 1$$

which is a potential range for $y$

Answer 2

Hint $$-1\leq sin(x) \leq1$$ now log isnt defined for negative values of so $0< y< 1$ for log and accordingly $x$

Answer 3

Taking from your last line: $y = \dfrac{2 - 4\sin^2 x + 4\sin x - 1}{2}= \dfrac{-4\sin^2x + 4\sin x + 1}{2} = f(t), - 1\leq t \leq 1, t = \sin x$. You can find the min and max of $f(t)$ and it will give you the range for $y$.