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Write equivalent equations in the form of inverse functions for
a.)x=y+cos è

(can you show how you would solve)

x= y+ cos è
cos è = x-y
theta = cos^-1(x-y)

cos(y) = x^2
y = Cos^-1(x^2)

  • Trigonometry -

    They are both correct!

  • Trigonometry -

    in a) is è a variable or a constant.
    If it is a variable then we don't find an "inverse"
    (the inverse is found by interchanging the x and y variables in a 2 variable relation)

    in b)
    original : cosy = x^2
    inverse: cosx = y^2

    solving this for y:
    y = ±√cosx,
    solving this for x:
    x = cos^-1 (y^2) , where -1 < y < 1

  • Trigonometry -

    Reiny, I interpret "inverse" being inverse trigonometric function, since he is doing a trigonometry course. However, I stand to be corrected!

  • Trigonometry -

    You are right.
    Funny how one's mind can get stuck along one track, and other possibilities get blocked.

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