Why is $arcsin (sin (12 π))$ $arcsin(sin(12pi))$ = 0 and not 0 and $2 π$ $2pi$ ?

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Why is $arcsin (sin (12 π))$ $arcsin(sin(12pi))$ = 0 and not 0 and $2 π$ $2pi$ ?

I thought it would be equal to 0 and $2 π$ $2pi$ , but the answers said 0 only.

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Answer 1 · 2017-12-04T09:05:16

arcsine is a function so for each input there is only one output.

Explanation:

The range of arcsine is $- \frac{π}{2} \leq y \leq \frac{π}{2}$ $-pi/2<=y<=pi/2$ , each $x$ $x$ -value in the domain is associated with only one $y$ $y$ -value. The only value in the range of arcsine that is associated with $x = 0$ $x=0$ , which is the value of $sin (12 π)$ $sin(12pi)$ , is $y = 0$ $y=0$ .

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Why is $arcsin (sin (12 π))$ $arcsin(sin(12pi))$ = 0 and not 0 and $2 π$ $2pi$ ?

I thought it would be equal to 0 and $2 π$ $2pi$ , but the answers said 0 only.

1 Answer

Explanation:

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Why is arcsin(sin(12pi))arcsin(sin(12π))arcsin(sin(12pi)) = 0 and not 0 and 2pi2π2pi?

I thought it would be equal to 0 and 2pi2π2pi, but the answers said 0 only.

1 Answer

Explanation:

Related questions

Impact of this question

Why is $arcsin (sin (12 π))$ $arcsin(sin(12pi))$ = 0 and not 0 and $2 π$ $2pi$ ?

I thought it would be equal to 0 and $2 π$ $2pi$ , but the answers said 0 only.