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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