Find at least one solution to the following equation: sin(x
2 − 1)
1 − sin(x
2 − 1) = sin(x) + sin2
(x) + sin3
(x) + sin4
(x) + · · ·

For |x| < 1, we have

We have |sin(x)| ≤ 1, with equality when x = ± π/2. For either of these, the right side of the equation does not converge, since it's either an infinite sum of 1s or an infinite alternating sum of 1 and -1.

So for |sin(x)| < 1, we have

so the equation is equivalent to

One obvious set of solutions occurs when x = x² - 1 :

x² - x - 1 = 0   →   x = (1 ± √(5))/2

i.e. the golden ratio φ ≈ 1.618 and 1 - φ ≈ 0.618.

step-by-step explanation:

where is the questions

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.