Posted by **Abbey** on Sunday, January 8, 2012 at 10:34am.

solve for x in the following equations:

1) sin(arcsin x) = 1

2) 2arcsin x = 1

3) cos(arccos x) = 1/3

- trig -
**Reiny**, Sunday, January 8, 2012 at 10:57am
think of the definition of inverse trig functions

arcsin x is the angle Ø so that sinØ = x

so let arcsin x = Ø

then we have sin Ø = 1

so if sinØ = 1 and sinØ = x

x = 1

(we could see that right away since something like sin(arcsin 30°) = 30° )

2.

2 arcsin x = 1

arcsin x = 1/2

I know that sin π/6 = 1/2 or sin 30° = 1/2

so x = π/6 ( or 5π/6 , since the sine is also positive in quadrant II )

3. Just like in #1, x = 1/3

- trig -
**Abbey**, Sunday, January 8, 2012 at 11:04am
Okay thank you for the help on one and three. In my book for number two it says the answer is sin1/2 which is approx. 0.479. I'm not sure how that answer comes about.

- correction - trig -
**Reiny**, Sunday, January 8, 2012 at 11:12am
My apologies

I solved as if it were 2sin x = 1

Don't know how I missed the "arc" since the whole question deals with it.

Anyway:

2 arcsinx = 1

arcsin x = 1/2

which by definition means

x = sin 1/2

so set your calculator to radians, and take sin .5

you will get .479...

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