# How do I simplify arcsin (sin 6 pi) given the interval 0 ≤ theta < 2pi

## since sin(?) and arcsin(? are inverses of each other,

arcsin( sin (6π) ) = 6π , but that is beyond your stated boundary.

so arcsin(sin 6π)
= arcsin 0
= 0 , π , 2π

## Ah, you've stumbled upon a classic math dilemma, my friend! Now, let's dive into the world of circular functions and unleash some clownery!

If we have arcsin(sin(6π)), the first step is to understand that arcsin and sin are not mortal enemies, but rather frenemies. Although sin(6π) equals 0, the arcsin function is a little peculiar. It's like a sneaky joker that only gives you answers between -π/2 and π/2, nothing more.

Since 6π falls outside this range, we have to employ some mathematical magic. To find the equivalent angle within the valid range, we use the concept of period. The sine function has a period of 2π, which means it repeats itself every 2π units.

To simplify, we need to find an angle that's within the range of -π/2 to π/2, but also equivalent to 6π. By subtracting multiples of 2π from 6π, we can bring it into the desired range. In this case, we subtract 3(2π) from 6π, leaving us with 0 as the equivalent angle. Therefore, simplifying arcsin(sin(6π)) gives us... drumroll, please... 0!

So, no clowning around, the simplified answer is 0. Keep in mind, though, this approach only works because the function sin(6π) happened to simplify to 0. In other cases, additional steps may be needed.

## To simplify arcsin(sin 6π) in the interval 0 ≤ θ < 2π, you can use the fact that arcsin(sinθ) = θ if θ is in the range -π/2 ≤ θ ≤ π/2.

Since 6π is outside the range -π/2 ≤ θ ≤ π/2, we need to bring it within that range. To do this, we can subtract or add multiples of 2π until we get an angle in the desired range.

In this case, we have 6π, which is equal to 3π + 3π. Since 3π is equivalent to π in the interval 0 ≤ θ < 2π, we can simplify 6π as follows:

6π = 3π + 3π = π + π + π = 3π

Now, we can take the arcsin of sin(3π):

arcsin(sin 3π) = arcsin(0) = 0

So, the simplified form of arcsin(sin 6π) in the interval 0 ≤ θ < 2π is 0.

## To simplify arcsin(sin(6π)) within the given interval 0 ≤ θ < 2π, you need to utilize the properties of the trigonometric functions.

The arcsin function returns the angle whose sine is a given value. However, in this case, sin(6π) is already given as an input, making it simpler.

By using the periodicity of the sine function, you can rewrite sin(6π) as sin(4π + 2π). Since the sine function has a period of 2π, sin(4π + 2π) is equivalent to sin(2π).

Now, sin(2π) is known to be equal to 0. This is because the sine function repeats its values every 2π, and at 2π, it reaches its first complete cycle where it starts from 0 again.

Therefore, the simplified expression arcsin(sin(6π)) within the interval 0 ≤ θ < 2π is arcsin(0), which is equal to 0 radians or 0 degrees.

In summary:
arcsin(sin(6π)) = arcsin(sin(4π + 2π)) = arcsin(sin(2π)) = arcsin(0) = 0 radians or 0 degrees.