# Calculate arcsin(sin√5)

1. Aug 2, 2010

### TsAmE

1. The problem statement, all variables and given/known data

Calculate arcsin(sin√5).

2. Relevant equations

None.

3. 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

2. Aug 2, 2010

### Staff: Mentor

Re: arcsin(sin√5)

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].

3. Aug 2, 2010

### The Chaz

Re: arcsin(sin√5)

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))?

4. Aug 2, 2010

### gomunkul51

Re: arcsin(sin√5)

you need to calculate arcsin(sin√5) ?

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

5. Aug 2, 2010

### Staff: Mentor

Re: arcsin(sin√5)

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

6. Aug 2, 2010

### gomunkul51

Re: arcsin(sin√5)

yup. my mistake.

7. Aug 2, 2010

### eumyang

Re: arcsin(sin√5)

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

8. Aug 2, 2010

### Bohrok

Re: arcsin(sin√5)