Question

Prove the following result: $\dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\left( 3-2\sqrt{2} \right)$.

Answer 1

Hint: We start solving the problem by using the properties $\cos \left( {{90}^{\circ }}+\theta \right)=-\sin \theta $ for the terms given in numerator and denominator. We then substitute the value of sine terms in numerator and denominator. We then multiply the numerator and denominator with the conjugate surd of the denominator to get the integer in the denominator. We then make the necessary calculations to prove the required result in the problem.

Complete step-by-step answer:
According to the problem, we need to prove the given result: $\dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\left( 3-2\sqrt{2} \right)$.
Let us first solve L.H.S (Left Hand Side) to prove that it is equal to R.H.S (Right Hand Side).
Now, we consider $\dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}$.
$\Rightarrow \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{\cos \left( {{90}^{\circ }}+{{45}^{\circ }} \right)-\cos \left( {{90}^{\circ }}+{{30}^{\circ }} \right)}{\cos \left( {{90}^{\circ }}+{{45}^{\circ }} \right)+\cos \left( {{90}^{\circ }}+{{30}^{\circ }} \right)}$ ---(1).
We know that $\cos \left( {{90}^{\circ }}+\theta \right)=-\sin \theta $. We use this result on equation (1).
So, we get $\dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{-\sin \left( {{45}^{\circ }} \right)-\left( -\sin \left( {{30}^{\circ }} \right) \right)}{-\sin \left( {{45}^{\circ }} \right)+\left( -\sin \left( {{30}^{\circ }} \right) \right)}$.
$\Rightarrow \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{-\sin \left( {{45}^{\circ }} \right)+\sin \left( {{30}^{\circ }} \right)}{-\sin \left( {{45}^{\circ }} \right)-\sin \left( {{30}^{\circ }} \right)}$ ---(2).
We know that $\sin \left( {{45}^{\circ }} \right)=\dfrac{1}{\sqrt{2}}$ and $\sin \left( {{30}^{\circ }} \right)=\dfrac{1}{2}$. We use these results in equation (2).
So, we get $\dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{-\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}}{-\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}}$.
$\Rightarrow \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{\dfrac{-\sqrt{2}+1}{2}}{\dfrac{-\sqrt{2}-1}{2}}$.
$\Rightarrow \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{-\sqrt{2}+1}{-\sqrt{2}-1}$.
$\Rightarrow \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}$ ---(3).
Let us multiply numerator and denominator in equation (3) with $\sqrt{2}-1$.
$\Rightarrow \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\times \dfrac{\sqrt{2}-1}{\sqrt{2}-1}$.
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.
$\Rightarrow \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{{{\left( \sqrt{2}-1 \right)}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}-{{1}^{2}}}$.
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.
$\Rightarrow \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{{{\left( \sqrt{2} \right)}^{2}}+{{1}^{2}}-2\left( \sqrt{2} \right)\left( 1 \right)}{2-1}$.
$\Rightarrow \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=\dfrac{2+1-2\sqrt{2}}{1}$.
$\Rightarrow \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=3-2\sqrt{2}$.
$\Rightarrow $ L.H.S = R.H.S.
So, we have proved $\dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=3-2\sqrt{2}$.

Note: We can also use the sum to product formulas $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and $\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$, but it will give the angles $127{{\dfrac{1}{2}}^{\circ }}$ and $17{{\dfrac{1}{2}}^{\circ }}$ which will be difficult to calculate. We can see that the problem contains a heavy amount of calculations, so we need to perform each step carefully. Whenever we get this type of problems, we try to convert the given angles in to the angles between ${{0}^{\circ }}$ and ${{90}^{\circ }}$ as we can use some of the standard values.