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Question: Find linear expression of sininversesinx for x belongs to [-2029π/2,-2025π/2]...

Find linear expression of sininversesinx for x belongs to [-2029π/2,-2025π/2]

Answer

sin⁻¹(sin x) = x + 1014π, x ∈ [-2029π/2, -2027π/2],

- x - 1013π,  x ∈ [-2027π/2, -2025π/2].
Explanation

Solution

Solution Explanation:

For any real xx, the identity

sin⁡−1(sin⁡x)=(−1)n(x−nπ)\sin^{-1}(\sin x) = (-1)^n (x - n\pi)

holds if

x∈[nπ−π2, nπ+π2].x \in \left[n\pi-\frac{\pi}{2},\, n\pi+\frac{\pi}{2}\right].

The given interval is

x∈[−2029π2,−2025π2],x\in\left[-\frac{2029\pi}{2},-\frac{2025\pi}{2}\right],

which is of length 2π2\pi. This interval splits naturally into two consecutive subintervals where the formula applies with different integers nn.

  1. First subinterval:

Let x∈[−2029π2,−2027π2]x\in\left[-\frac{2029\pi}{2},-\frac{2027\pi}{2}\right]. We choose nn such that

nπ−π2=−2029π2.n\pi-\frac{\pi}{2} = -\frac{2029\pi}{2}.

Multiply by 2:

2nπ−π=−2029π⟹2nπ=−2029π+π=−2028π.2n\pi - \pi = -2029\pi \quad\Longrightarrow\quad 2n\pi = -2029\pi + \pi = -2028\pi.

Hence,

n=−1014.n = -1014.

Since −1014-1014 is even, (−1)−1014=1(-1)^{-1014} = 1. Thus,

sin⁡−1(sin⁡x)=x−(−1014π)=x+1014πfor x∈[−2029π2,−2027π2].\sin^{-1}(\sin x)= x - (-1014\pi)= x + 1014\pi \quad \text{for } x\in\left[-\frac{2029\pi}{2},-\frac{2027\pi}{2}\right].
  1. Second subinterval:

Let x∈[−2027π2,−2025π2]x\in\left[-\frac{2027\pi}{2},-\frac{2025\pi}{2}\right]. Here, select nn such that

nπ+π2=−2025π2.n\pi+\frac{\pi}{2} = -\frac{2025\pi}{2}.

Multiply by 2:

2nπ+π=−2025π⟹2nπ=−2025π−π=−2026π,2n\pi + \pi = -2025\pi \quad\Longrightarrow\quad 2n\pi = -2025\pi - \pi = -2026\pi,

so,

n=−1013.n= -1013.

Since −1013-1013 is odd, (−1)−1013=−1(-1)^{-1013} = -1. Hence,

sin⁡−1(sin⁡x)=−(x−(−1013π))=−(x+1013π)=−x−1013πfor x∈[−2027π2,−2025π2].\sin^{-1}(\sin x)= -\left(x - (-1013\pi)\right)= -\left(x + 1013\pi\right)= -x - 1013\pi \quad \text{for } x\in\left[-\frac{2027\pi}{2},-\frac{2025\pi}{2}\right].