Question

Prove that $\dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}=3-2\sqrt{2}$.

Answer 1

For solving this type of question you should know about the trigonometric values of all trigonometric functions and should know to convert the angles according to the quadrant. Because after changing the values of the angles, the trigonometric function also changes according to quadrant changing.

Complete step by step answer:
In this question $\dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}$ is given and we have to prove it is equal to $3-2\sqrt{2}$. So, by the trigonometric formulas we convert the angles and determine the values. So, we have,
$\begin{aligned} & \cos \left( {{135}^{\circ }} \right)=\cos \left( 180-45 \right)=-\cos {{45}^{\circ }} \\\ & \because \cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}} \\\ & \therefore -\cos {{45}^{\circ }}=-\dfrac{1}{\sqrt{2}} \\\ \end{aligned}$
And the same goes for, so we will convert all the possible angles greater than 180 to make them less than 90. Therefore, we get
$\begin{aligned} & \cos \left( {{120}^{\circ }} \right)=\cos \left( 180-60 \right)=-\cos {{60}^{\circ }} \\\ & \because \cos {{60}^{\circ }}=-\dfrac{1}{2} \\\ & \therefore -\cos {{45}^{\circ }}=\dfrac{1}{2} \\\ \end{aligned}$
So, according to the question we will put the values in the given expression,
$\begin{aligned} & \dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}} \\\ & =\dfrac{-\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}}{-\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}} \\\ \end{aligned}$
After solving this equation by taking negative sign common from both numerator and denominator, we get,
$\dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}}$
By multiplying the denominator and the numerator by 2, we will get,
$\dfrac{\dfrac{2}{\sqrt{2}}-1}{\dfrac{2}{\sqrt{2}}+1}$
So, by multiplying above and below by $\left( \dfrac{2}{\sqrt{2}}-1 \right)$ .
Thus by multiplying we get,
$\dfrac{\left( \dfrac{2}{\sqrt{2}}-1 \right)}{\left( \dfrac{2}{\sqrt{2}}+1 \right)}\times \dfrac{\left( \dfrac{2}{\sqrt{2}}-1 \right)}{\left( \dfrac{2}{\sqrt{2}}-1 \right)}$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, so we can write the above as,
$\begin{aligned} & \dfrac{{{\left( \dfrac{2}{\sqrt{2}}-1 \right)}^{2}}}{{{\left( \dfrac{2}{\sqrt{2}} \right)}^{2}}-{{\left( 1 \right)}^{2}}} \\\ & =\dfrac{2+1-2\sqrt{2}}{2-1} \\\ & =3-2\sqrt{2} \\\ \end{aligned}$
So, it is proved that $\dfrac{\cos {{135}^{\circ }}-\cos {{120}^{\circ }}}{\cos {{135}^{\circ }}+\cos {{120}^{\circ }}}$ is equal to $3-2\sqrt{2}$ .

Note: We can also solve this question by the trigonometric formula, $\dfrac{\cos A-\cos B}{\cos A+\cos B}=\dfrac{-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)}{2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}$. And this is also a very useful and easy method for solving this question. But calculating and putting the values are very careful points in this question.