Hi I was just wondering how do you do inverse functions?
example: Sin^-1 (5pi/6)
Sin^-1 (3pi/2)
Why does most of the answers only have 1 answer and is out of "RANGE"?
Sin^-1 sqrt(3)/2
Has only 1 answer. Why is 2pi/3 out of the range?
The interval for the range of sin(x) is
[-1,1].
5Ï€/6=2.61... > 1
and similarly
3Ï€/2=4.71... >1
sin-1(sqrt(3)/2)
= π/3
To find the inverse function, you need to think of it as "undoing" what the original function did. In the case of trigonometric functions like sine, cosine, and tangent, these functions take an angle as input and produce a ratio as output. The inverse functions, denoted by sin^(-1), cos^(-1), and tan^(-1), take a ratio as input and produce an angle as output.
To find the value of sin^(-1)(x), you need to find the angle whose sine value is x. However, sine is a periodic function, meaning it repeats itself every 360 degrees (or 2π radians). So, there are infinitely many angles with the same sine value. To limit the output to a single value, we typically restrict the range of the inverse function to [-π/2, π/2] or [-90°, 90°], which corresponds to the principal branch.
For example, if you have sin^(-1)(1/2), you are looking for the angle whose sine is 1/2. Since sine is positive for angles in the first and second quadrants, we know that the angle must be between -90° and 90°. The angle that satisfies this condition is 30° (or π/6 radians). Therefore, sin^(-1)(1/2) = 30°.
In your examples:
1) sin^(-1)(5π/6) is asking for the angle whose sine is 5π/6. Since sine is positive for angles in the first and second quadrants, you need to find an angle between -90° and 90° whose sine is 5π/6. However, 5π/6 does not fall within this range, so the answer is out of range.
2) sin^(-1)(3π/2) is asking for the angle whose sine is 3π/2. Again, since sine is positive for angles in the first and second quadrants, you need to find an angle between -90° and 90° whose sine is 3π/2. However, 3π/2 is not a valid sine value. Hence, the answer is out of range.
In cases where the input falls outside the restricted range, the answer is considered out of range and is not a valid angle for the inverse function.