What is the result of calculating arcsin(sin√5)?

[TsAmE]

Homework Statement

Calculate arcsin(sin√5).

Homework Equations

None.

The Attempt at a Solution

f(1/2) = sin(1/2)
= π / 6

f'(x) = cosx

f'(1/2) = cos(1/2)
= π / 3

y = π / 6 + π / 3(x - 1/2)
= π / 3 x

f(√5) = π / 3 (√5)
= √5π / 3

I tried using linear approximation, but didnt get the correct answer of π - √5

 

[Mark44]

arcsin(sin x) = x, PROVIDED that x is in the interval [-pi/2, pi/2]. You need to find a number x such that sin(x) = sin(sqrt(5)), where x is in [-pi/2, pi/2].

 

[The Chaz]

TsAmE, I might add that Calculus *methods* are not appropriate for this problem (even though it might appear in a calc book/course).
This is more about, as Mark44 has mentioned, coterminal angles and the domain of inverse trig functions.

Some people have an easier time working with degrees at first. Could you calculate
arcsin(sin(113degrees))?

 

[gomunkul51]

you need to calculate arcsin(sin√5) ?

arcsin and sin are inverse functions of each other, thus:
arcsin(sin(x)) = x :)

 

[Mark44]

you need to calculate arcsin(sin√5) ?

arcsin and sin are inverse functions of each other, thus:
arcsin(sin(x)) = x :)

This is not true in general, and definitely not true in this case arcsin(sin($\sqrt{5}$)) $\neq$ $\sqrt{5}$. See post #2.

 

[gomunkul51]

This is not true in general, and definitely not true in this case arcsin(sin($\sqrt{5}$)) $\neq$ $\sqrt{5}$. See post #2.

yup. my mistake.

 

[eumyang]

I don't know how much is assumed that you know in terms of numbers, but I remember that
$$\sqrt{5} \approx 2.236$$

... and I remember the radian equivalent of the angles for each quadrant (for example, the angles in Q1 go from 0 to ~1.571 rad), so I could figure out what quadrant $$\sqrt{5}\,rad$$ is.

Also, it would be helpful to know one of the trig identities that deal with symmetry. (I hope that wasn't too many hints.)69

 

[Bohrok]

Homework Statement

Calculate arcsin(sin√5).

I tried using linear approximation, but didnt get the correct answer of π - √5

Use the identity sin(x) = sin([itex]\pi[/tex] - x)

 

[TsAmE]

Use the identity sin(x) = sin([itex]\pi[/tex] - x)

If that is the case then:

arcsin(sin√5) = arc(sin(π - √5)), but π - √5 is still in the 2nd quadrant which doesn't satisfy arc(sinx) = x [-π/2, π/2].

 

[Mentallic]

Not quite. By intuition, $\pi$ is a bit more than 3 and $\sqrt{5}$ is a bit more than 2, so $\pi-\sqrt{5}$ is approximately 1. But $\pi/2$ is a number a bit more than 3 and halved, so a bit more than 1.5. Obviously $\pi-\sqrt{5}<\pi/2$ so it is in the first quadrant.