we used sin⁻¹ to find the angle whose sine is 0.5.
Vi brugte sin⁻¹ til at finde vinklen, hvis sinus er 0,5.
the calculator showed the result of sin⁻¹(√3/2) was π/3.
Regneren viste, at resultatet af sin⁻¹(√3/2) var π/3.
to solve for θ, we applied the sin⁻¹ function to both sides.
For at løse for θ anvendte vi sin⁻¹-funktionen på begge sider.
the range of sin⁻¹ is restricted to [-π/2, π/2] for accurate results.
Definitionsmængden for sin⁻¹ er begrænset til [-π/2, π/2] for at få præcise resultater.
we need to ensure the argument of sin⁻¹ is within the domain.
Vi skal sikre, at argumentet for sin⁻¹ ligger inden for definitionsmængden.
the inverse sine, or sin⁻¹, is a trigonometric function.
Den inverse sinus, eller sin⁻¹, er en trigonometrisk funktion.
using sin⁻¹ on -1 gives us an angle of -π/2.
At bruge sin⁻¹ på -1 giver os en vinkel på -π/2.
the problem required finding the angle whose sine was 1/2 using sin⁻¹.
Problemet krævede, at vi fandt vinklen, hvis sinus var 1/2, ved brug af sin⁻¹.
we can find the reference angle using sin⁻¹ and the quadrant.
Vi kan finde den referencevinkel ved at bruge sin⁻¹ og kvadrantet.
the sin⁻¹ of 0 is 0 radians.
sin⁻¹ af 0 er 0 radianer.
the value of sin⁻¹(0.866) is approximately π/3.
Værdien af sin⁻¹(0,866) er omtrent π/3.
we used sin⁻¹ to find the angle whose sine is 0.5.
Vi brugte sin⁻¹ til at finde vinklen, hvis sinus er 0,5.
the calculator showed the result of sin⁻¹(√3/2) was π/3.
Regneren viste, at resultatet af sin⁻¹(√3/2) var π/3.
to solve for θ, we applied the sin⁻¹ function to both sides.
For at løse for θ anvendte vi sin⁻¹-funktionen på begge sider.
the range of sin⁻¹ is restricted to [-π/2, π/2] for accurate results.
Definitionsmængden for sin⁻¹ er begrænset til [-π/2, π/2] for at få præcise resultater.
we need to ensure the argument of sin⁻¹ is within the domain.
Vi skal sikre, at argumentet for sin⁻¹ ligger inden for definitionsmængden.
the inverse sine, or sin⁻¹, is a trigonometric function.
Den inverse sinus, eller sin⁻¹, er en trigonometrisk funktion.
using sin⁻¹ on -1 gives us an angle of -π/2.
At bruge sin⁻¹ på -1 giver os en vinkel på -π/2.
the problem required finding the angle whose sine was 1/2 using sin⁻¹.
Problemet krævede, at vi fandt vinklen, hvis sinus var 1/2, ved brug af sin⁻¹.
we can find the reference angle using sin⁻¹ and the quadrant.
Vi kan finde den referencevinkel ved at bruge sin⁻¹ og kvadrantet.
the sin⁻¹ of 0 is 0 radians.
sin⁻¹ af 0 er 0 radianer.
the value of sin⁻¹(0.866) is approximately π/3.
Værdien af sin⁻¹(0,866) er omtrent π/3.
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