Question

evaluate: arcsin (sin 2)​

Answer 1

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The range of arcsin is − π 2 ≤ θ ≤ π 2 . 2 π 2 so it lies outside the range.

The range of arcsin is − π 2 ≤ θ ≤ π 2 . 2 π 2 so it lies outside the range.Use sin ( π − θ ) = sin θ

The range of arcsin is − π 2 ≤ θ ≤ π 2 . 2 π 2 so it lies outside the range.Use sin ( π − θ ) = sin θ

The range of arcsin is − π 2 ≤ θ ≤ π 2 . 2 π 2 so it lies outside the range.Use sin ( π − θ ) = sin θ Then sin 2 = sin ( π − 2 )

The range of arcsin is − π 2 ≤ θ ≤ π 2 . 2 π 2 so it lies outside the range.Use sin ( π − θ ) = sin θ Then sin 2 = sin ( π − 2 )

The range of arcsin is − π 2 ≤ θ ≤ π 2 . 2 π 2 so it lies outside the range.Use sin ( π − θ ) = sin θ Then sin 2 = sin ( π − 2 ) π − 2 ≅ 1.14

The range of arcsin is − π 2 ≤ θ ≤ π 2 . 2 π 2 so it lies outside the range.Use sin ( π − θ ) = sin θ Then sin 2 = sin ( π − 2 ) π − 2 ≅ 1.14 so 0 < π − 2 < π 2

The range of arcsin is − π 2 ≤ θ ≤ π 2 . 2 π 2 so it lies outside the range.Use sin ( π − θ ) = sin θ Then sin 2 = sin ( π − 2 ) π − 2 ≅ 1.14 so 0 < π − 2 < π 2

The range of arcsin is − π 2 ≤ θ ≤ π 2 . 2 π 2 so it lies outside the range.Use sin ( π − θ ) = sin θ Then sin 2 = sin ( π − 2 ) π − 2 ≅ 1.14 so 0 < π − 2 < π 2 So arcsin ( sin ( 2 ) ) = π − 2

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Answer 2

the range of arcsin (x) is [-π/2,π/2]

Arcsin(sinx)=x if |x|≤π/2 since x=2>π/2 ..(π/2=1.57..) arcsin(sin2)=π-2