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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.

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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.

$$\frac{3-\sqrt{3}}{3+\sqrt3}=\frac{3-\sqrt3}{3+\sqrt3}\cdot\frac{3-\sqrt3}{3-\sqrt3}=\frac{9+3-6\sqrt3}{9-3}=2-\sqrt3$$

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  • $\begingroup$ what does the '.' between the two fractions mean? Are you multiplying the numerator and denominator by 3 - √3 ? $\endgroup$ – A.Mahony May 25 '17 at 12:40
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    $\begingroup$ The $\cdot$ just means the same thing as $\times$, so yes, multiplication $\endgroup$ – John Doe May 25 '17 at 12:41
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    $\begingroup$ @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. $\endgroup$ – origimbo May 25 '17 at 14:27
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You can go further: \begin{align}\frac{3(3-\sqrt3)}{3(3+\sqrt3)}&=\frac{3-\sqrt3}{3+\sqrt3}\\\\ &=\frac{9-6\sqrt3+3}{3^2-3}\\\\ &=\frac{12-6\sqrt3}6\\\\ &=2-\sqrt3\end{align}

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Here you go,
$$\begin{align}\tan \left( {{{45}^0} - {{30}^0}} \right) &= \frac{{\tan {{45}^0} - \tan {{30}^0}}}{{1 + \tan {{45}^0}\tan {{30}^0}}}\\\\ &= \frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \left( 1 \right)\left( {\frac{1}{{\sqrt 3 }}} \right)}}\\\\ &= \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\\ \\ &= \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}\\\\ &= \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - 1}}\\\\ &= \frac{{3 + 1 - 2\sqrt 3 }}{{3 - 1}}\\\\ &= \frac{{4 - 2\sqrt 3 }}{2}\\\\ &= 2 - \sqrt 3 \end{align} $$

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