Question

Find the value of $\tan \left( {{{15}^ \circ }} \right)$ using half angle formula.

Answer 1

The given problem can be solved by using the half angle formula of tangent. Use of the half angle formula helps us to convert $\tan \left( {2\theta } \right)$ to an expression consisting of $\tan \left( \theta \right)$ .

Half angle formula for tangent is: $\tan \left( {2x} \right) = \dfrac{{2\tan \left( x \right)}}{{1 - {{\tan }^2}\left( x \right)}}$ . This formula can be used to find a quadratic equation in $\tan \left( {{{15}^ \circ }} \right)$ as we know the value of $\tan \left( {{{30}^ \circ }} \right)$ .

Complete step by step solution:
The given problem requires us to find the value of $\tan \left( {{{15}^ \circ }} \right)$ using half angle formula. The half angle formula for tangent is: $\tan \left( {2x} \right) = \dfrac{{2\tan \left( x \right)}}{{1 - {{\tan }^2}\left( x \right)}}$ .

So, considering $x = {15^ \circ }$, we get to a quadratic equation in $\tan \left( {{{15}^ \circ }} \right)$ using the formula.

So, $\tan \left( {2 \times {{15}^ \circ }} \right) = \dfrac{{2\tan \left( {{{15}^ \circ }} \right)}}{{1 - {{\tan }^2}\left( {{{15}^ \circ }} \right)}}$ .
$= \tan \left( {{{30}^ \circ }} \right) = \dfrac{{2\tan \left( {{{15}^ \circ }} \right)}}{{1 - {{\tan }^2}\left( {{{15}^ \circ }} \right)}}$

Now, we know that the value of $\tan \left( {{{30}^ \circ }} \right)$ is $\dfrac{1}{{\sqrt 3 }}$ .

So, we get the equation as:
$= \dfrac{1}{{\sqrt 3 }} = \dfrac{{2\tan \left( {{{15}^ \circ }} \right)}}{{1 - {{\tan }^2}\left( {{{15}^ \circ }} \right)}}$

Cross multiplying the fractions and simplifying the calculations, we get,
$= 1 - {\tan ^2}\left( {{{15}^ \circ }} \right) = 2\sqrt 3 \tan \left( {{{15}^ \circ }} \right)$

Rearranging the equation into standard form of quadratic equation $a{x^2} + bx + c = 0$, we get,
$= {\tan ^2}\left( {{{15}^ \circ }} \right) + 2\sqrt 3 \tan \left( {{{15}^ \circ }} \right) - 1 = 0$

Now, the above quadratic equation can be solved by various methods such as: Completing the square method, Splitting the middle term method, factorisation method and using the quadratic formula.

We would use a quadratic formula to solve the above equation as this method involves minimal calculations. So, Let $\tan \left( {{{15}^ \circ }} \right) = y$. Then, the equation becomes ${y^2} + 2\sqrt 3 y - 1 = 0$ .

Comparing the equation ${y^2} + 2\sqrt 3 y - 1 = 0$ with the standard form of quadratic equation, $a{x^2} + bx + c = 0$, we get, $a = 1$, $b = 2\sqrt 3$ and $c = - 1$ .

So, Using quadratic formula,
$y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Putting the values,
$= \dfrac{{ - 2\sqrt 3 \pm \sqrt {{{\left( {2\sqrt 3 } \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}$
$= \dfrac{{ - 2\sqrt 3 \pm \sqrt {12 + 4} }}{2}$

Simplifying further,
$= \dfrac{{ - 2\sqrt 3 \pm 4}}{2}$
$= \left( { - \sqrt 3 \pm 2} \right)$
So, either $y = \left( { - \sqrt 3 + 2} \right)$ or $y = \left( { - \sqrt 3 - 2} \right)$ .

Substituting back $\tan \left( {{{15}^ \circ }} \right) = y$, we get,
either $\tan \left( {{{15}^ \circ }} \right) = \left( { - \sqrt 3 + 2} \right)$ or $\tan \left( {{{15}^ \circ }} \right) = \left( { - \sqrt 3 - 2} \right)$ .

Since, the angle is acute. Hence, it would lie in the first quadrant. So, $\tan \left( {{{15}^ \circ }} \right)$ must be positive.

Hence, the value of$\tan \left( {{{15}^ \circ }} \right) = \left( { - \sqrt 3 + 2} \right)$ .

Note: The above question can also be solved by using compound angle formulae instead of half angle formulae such as $\tan (A + B) = \left( {\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}} \right)$ . This method can also be used to get to the correct answer of the given problem.