Here sin-1 doesn't mean 1/sin, it means the inverse function. As the sine function has an angle as an argument and returns a trigonometric value, the inverse sine function uses a trigonometric value as a parameter and returns the angle, associated with that value. This also works for the other trigonometric functions, not only sine. In the video, is it clarified that the value returned by the inverse sin is between - 90 and 90 degrees, so only one degree value for possible trigonometric value (from 0 to 1 including).
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Should we also include negative values as part of the solution to the second part?
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1/sin(0,5)=x° 1sin(0,5)÷x°=1; x°= 114,59
siny°=1/2 siny°/0,5=1
y°=30 x°/y°=114,59/30=3,81
Here sin-1 doesn't mean 1/sin, it means the inverse function. As the sine function has an angle as an argument and returns a trigonometric value, the inverse sine function uses a trigonometric value as a parameter and returns the angle, associated with that value. This also works for the other trigonometric functions, not only sine. In the video, is it clarified that the value returned by the inverse sin is between - 90 and 90 degrees, so only one degree value for possible trigonometric value (from 0 to 1 including).