Find the angle $\theta $ whose cosine is equal to its tangent
- $\theta =\sin^{-1}\left ( 2\sin 18^{0} \right )$
- $\theta =\sin^{-1}\left ( 2\sin 36^{0} \right )$
- $\theta =\sin^{-1}\left ( 3\sin 18^{0} \right )$
- $\theta =\sin^{-1}\left ( 2\sin 54^{0} \right )$
We need cos(theta) = tan(theta) = sin(theta)/cos(theta), so cos^2(theta) = sin(theta). This is 1 - sin^2(theta) = sin(theta), or sin^2(theta) + sin(theta) - 1 = 0. Using the quadratic formula, sin(theta) = (-1 + sqrt(5))/2. This value is equal to 2 * sin(18 degrees).
Set up the equation based on the given condition, which is cos theta equals tan theta. Substituting tan theta as sin theta divided by cos theta gives cos squared theta equals sin theta. Using the Pythagorean identity, this becomes 1 minus sin squared theta equals sin theta, resulting in the quadratic equation sin squared theta plus sin theta minus 1 equals 0. Solving this quadratic equation for sin theta yields positive one half times negative 1 plus the square root of 5, which is standardly known to equal 2 times sin 18 degrees by the golden ratio properties. Therefore, theta equals arcsine of 2 times sin 18 degrees.