# What is x if arccos x-arcsin x=pie/6?

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You should use the following notations for `arccos x` and arcsin x, such that:

`arccos x = alpha`

`arcsin x = theta`

You need to take the sine function both sides, such that:

`sin(alpha - theta) = sin(pi/6)`

Expanding the left side yields:

`sin alpha*cos theta - sin theta*cos alpha = 1/2`

You need to use the following trigonometric identities, such that:

`sin(arcsin x) = cos(arccos x) = x`

`sin(arccos x) = cos(arcsin x) = sqrt(1 - x^2)`

Reasoning by analogy yields:

`sqrt(1 - x^2)*sqrt(1 - x^2) - x*x = 1/2`

`(1 - x^2) - x^2 = 1/2 => 1 - 2x^2 = 1/2 => -2x^2 = 1/2 - 1`

`-2x^2 = -1/2 => 2x^2 = 1/2 => x^2 = 1/4 => x_(1,2) = +-sqrt(1/4) => x_(1,2) = +-1/2`

Testing the values in equation yields:

`x = 1/2 => arccos(1/2) - arcsin(1/2) = pi/3 - pi/6 = (2pi - pi)/6 = pi/6` valid

`x = -1/2 => arccos(-1/2) - arcsin(-1/2) = pi - pi/3 + pi/6 = 2pi/3 + pi/6 = (5pi/6) != pi/6`

**Hence, evaluating the solutions to the given equation yields **`x = 1/2.`

**Sources:**

Let

`arcsin(x)=alpha`

`x=sin(alpha)`

`sin(alpha)=x`

`cos(pi/2-alpha)=x`

`pi/2-alpha=arccos(x)`

Thus

`arccos(x)-arcsin(x)=pi/6`

`pi/2-alpha-alpha=pi/6`

`pi/2-pi/6=2alpha`

`(2pi)/6=2alpha`

`alpha=pi/6`

`Thus`

`arcsin(x)=pi/6`

`x=sin(pi/6)`

`x=1/2`

**Note:**

`arcsin(x)+arccos(x)=pi/2`