Find the Exact Value tan(arccos(-12/13))

Problem

tan⁡arccos⁡(−1213) the tangent of the arc cosine of open paren negative 12 over 13 close paren

Solution

  1. Identify the angle θ theta defined by the inverse cosine function such that θ=arccos⁡(−1213) theta equals the arc cosine of open paren negative 12 over 13 close paren

  2. Determine the range of the inverse cosine function, which is [0,π] open bracket 0 comma pi close bracket Since the argument −1213 negative 12 over 13 is negative, θ theta must be in the second quadrant (Quadrant II).

  3. Use the definition of cosine in terms of x x y y and r r where cos⁡θ=xr cosine theta equals x over r Here, x=−12 x equals negative 12 and r=13 r equals 13

  4. Calculate the value of y y using the Pythagorean identity x2+y2=r2 x squared plus y squared equals r squared

(−12)2+y2=132 open paren negative 12 close paren squared plus y squared equals 13 squared

144+y2=169 144 plus y squared equals 169

y2=25 y squared equals 25

  1. Select the positive root y=5 y equals 5 because the sine (and thus y y is positive in Quadrant II.

  2. Apply the definition of tangent, which is tan⁡θ=yx tangent theta equals y over x

tan⁡θ=5−12 tangent theta equals 5 over negative 12

Final Answer

tan⁡arccos⁡(−1213)=−512 the tangent of the arc cosine of open paren negative 12 over 13 close paren equals negative 5 over 12

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