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.

last option

step-by-step explanation:

given that δdef is right and isoceles,

assuming that side df is the hypotenuse means that de an ef are the other 2 sides that intersect at 90° and are of equal length.

because they are of equal length, we can say that they are congruent

because they intersect at 90°, their slopes must be opposite reciprocals.

only the last option satisfies these conditions.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: