# trig

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solve for x in the following equations:

1) sin(arcsin x) = 1

2) 2arcsin x = 1

3) cos(arccos x) = 1/3

• trig -

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 -

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 -

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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