# Determine exact values using angle sum and difference identity (trig)

Ok so we have just learnt the basic angle sum and difference identities:

$$\begin{array}{l}\cos \left( {A \pm B} \right) = \cos A\cos B \mp \sin A\sin B\\\sin \left( {A \pm B} \right) = \sin A\cos B \pm \cos A\sin B\\\tan \left( {A \pm B} \right) = \frac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}\end{array}$$

The question i have been given is determine the exact value of $\tan {15^0}$ degrees. So i assume you use the tan identity, sub in

$$\tan \left( {{{45}^0} - {{30}^0}} \right) = \frac{{\tan {{45}^0} - \tan {{30}^0}}}{{1 + \tan {{45}^0}\tan 30}}$$

After simplifying it down as far as possible i end up with:

$$\frac{{\left( {3\left( {3 - \sqrt 3 } \right)} \right)}}{{\left( {3\left( {3 + \sqrt 3 } \right)} \right)}}$$

However the correct answer is: ${2 - \sqrt 3 }$

I'm not sure if iv'e done something wrong or are completely off track, but could someone please explain (as simply as possible) how to achieve this answer.

You haven't done anything wrong, you just haven't finished yet - it is good practice to not leave irrational numbers ($\sqrt3$) in the denominator.
$$\frac{3-\sqrt{3}}{3+\sqrt3}=\frac{3-\sqrt3}{3+\sqrt3}\cdot\frac{3-\sqrt3}{3-\sqrt3}=\frac{9+3-6\sqrt3}{9-3}=2-\sqrt3$$
• The $\cdot$ just means the same thing as $\times$, so yes, multiplication – John Doe May 25 '17 at 12:41
• @A.Mahony Just in case it's not clear, (I know it probably is) the core identity being used here is $(a+b)\times(a-b)=a^2-b^2$. This provides a fairly general way to clean up denominators containing the sum of an integer and a surd. – origimbo May 25 '17 at 14:27
\begin{align}\tan \left( {{{45}^0} - {{30}^0}} \right) &= \frac{{\tan {{45}^0} - \tan {{30}^0}}}{{1 + \tan {{45}^0}\tan {{30}^0}}}\\\\ &= \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \left( 1 \right)\left( {\frac{1}{{\sqrt 3 }}} \right)}}\\\\ &= \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\\ \\ &= \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}\\\\ &= \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - 1}}\\\\ &= \frac{{3 + 1 - 2\sqrt 3 }}{{3 - 1}}\\\\ &= \frac{{4 - 2\sqrt 3 }}{2}\\\\ &= 2 - \sqrt 3 \end{align}