# what is `tan(cos^(-)1 (1/4))` ?

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Determine `tan(cos^(-1)1/4)` :

The arccos is defined on the interval from 0 to `pi` ; since the cos is positive in the first quadrant we look in the first quadrant.

Draw a right triangle in the first quadrant: the side along the x-axis has length 1, the right angle is located at x=1, and the length of the hypotenuse is 4. Then by the pythagorean theorem the other leg has length `sqrt(15)` .

The angle we are interested in has its vertex at the origin; call it `alpha` . Then `tan alpha=("side" "opposite")/("side" "adjacent")=(sqrt(15))/1=sqrt(15)`

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`tan(cos^(-1)1/4)=sqrt(15)`

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**Sources:**

To visualize this, draw a right triangle. Label one of the non-right angles "x." You want cos x = 1/4, so label the adjacent side of angle x with a 1 and the hypotenuse with 4.

Using the Pythagorean theorem, the remaining side of the triangle is sqrt(15).

cos x = 1/4, so x = cos^-1 (1/4)

tan (cos^-1 (1/4) ) = tan (x) = sqrt(15) / 1 = sqrt(15)