Finding the Value of tan 18°

To find the value of tan 18°, we can use the values of sin 18° and cos 18°, as tan \(\theta = \frac{\sin \theta}{\cos \theta}\).

First, let's find the value of sin 18°. Let \(\theta = 18^\circ\). Then \(5\theta = 90^\circ\). We can write this as \(2\theta = 90^\circ - 3\theta\). Taking the sine of both sides, we get:

\(\sin(2\theta) = \sin(90^\circ - 3\theta)\)

Using the double angle formula for sine and the co-function identity, we have:

\(2\sin\theta\cos\theta = \cos(3\theta)\)

Now, using the triple angle formula for cosine, \(\cos(3\theta) = 4\cos^3\theta - 3\cos\theta\):

\(2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta\)

Since \(\cos 18^\circ \neq 0\), we can divide both sides by \(\cos\theta\):

\(2\sin\theta = 4\cos^2\theta - 3\)

Using the identity \(\cos^2\theta = 1 - \sin^2\theta\):

\(2\sin\theta = 4(1 - \sin^2\theta) - 3\)

\(2\sin\theta = 4 - 4\sin^2\theta - 3\)

\(2\sin\theta = 1 - 4\sin^2\theta\)

Rearranging the terms to form a quadratic equation in terms of \(\sin\theta\):

\(4\sin^2\theta + 2\sin\theta - 1 = 0\)

Let \(x = \sin\theta\). The equation becomes \(4x^2 + 2x - 1 = 0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(x = \frac{-2 \pm \sqrt{2^2 - 4(4)(-1)}}{2(4)}\)

\(x = \frac{-2 \pm \sqrt{4 + 16}}{8}\)

\(x = \frac{-2 \pm \sqrt{20}}{8}\)

\(x = \frac{-2 \pm 2\sqrt{5}}{8}\)

\(x = \frac{-1 \pm \sqrt{5}}{4}\)

Since \(\theta = 18^\circ\) is in the first quadrant, \(\sin 18^\circ\) must be positive. Therefore,

\(\sin 18^\circ = \frac{\sqrt{5} - 1}{4}\)

Next, let's find the value of cos 18°. We use the identity \(\cos^2\theta = 1 - \sin^2\theta\):

\(\cos^2 18^\circ = 1 - \left(\frac{\sqrt{5} - 1}{4}\right)^2\)

\(\cos^2 18^\circ = 1 - \frac{(\sqrt{5})^2 - 2\sqrt{5}(1) + 1^2}{16}\)

\(\cos^2 18^\circ = 1 - \frac{5 - 2\sqrt{5} + 1}{16}\)

\(\cos^2 18^\circ = 1 - \frac{6 - 2\sqrt{5}}{16}\)

\(\cos^2 18^\circ = 1 - \frac{3 - \sqrt{5}}{8}\)

\(\cos^2 18^\circ = \frac{8 - (3 - \sqrt{5})}{8}\)

\(\cos^2 18^\circ = \frac{8 - 3 + \sqrt{5}}{8}\)

\(\cos^2 18^\circ = \frac{5 + \sqrt{5}}{8}\)

Since 18° is in the first quadrant, cos 18° is positive. So,

\(\cos 18^\circ = \sqrt{\frac{5 + \sqrt{5}}{8}}\)

We can rewrite this expression:

\(\cos 18^\circ = \frac{\sqrt{5 + \sqrt{5}}}{\sqrt{8}}\)

Multiply numerator and denominator by \(\sqrt{2}\) to get a denominator of 4:

\(\cos 18^\circ = \frac{\sqrt{2}\sqrt{5 + \sqrt{5}}}{\sqrt{8}\sqrt{2}}\)

\(\cos 18^\circ = \frac{\sqrt{2(5 + \sqrt{5})}}{\sqrt{16}}\)

\(\cos 18^\circ = \frac{\sqrt{10 + 2\sqrt{5}}}{4}\)

Finally, we can find the value of tan 18°:

\(\tan 18^\circ = \frac{\sin 18^\circ}{\cos 18^\circ}\)

\(\tan 18^\circ = \frac{\frac{\sqrt{5} - 1}{4}}{\frac{\sqrt{10 + 2\sqrt{5}}}{4}}\)

Cancel out the denominator 4 from both numerator and denominator:

\(\tan 18^\circ = \frac{\sqrt{5} - 1}{\sqrt{10 + 2\sqrt{5}}}\)

This matches one of the given options.

Let's check the options provided:

• Option 1: \(\frac{{\sqrt 5 - 1}}{{\sqrt {10 + 2\sqrt 5 } }}\)
• Option 2: \(\frac{{\sqrt 5 - 1}}{{\sqrt {10 + \sqrt 5 } }}\)
• Option 3: \(\frac{{\sqrt {10 + 2\sqrt 5 } }}{{\sqrt 5 - 1}}\)
• Option 4: \(\frac{{\sqrt {10 + \sqrt 5 } }}{{\sqrt 5 - 1}}\)

Comparing our derived value with the options, the value of tan 18° is \(\frac{{\sqrt 5 - 1}}{{\sqrt {10 + 2\sqrt 5 } }}\).