Question

Evaluate the given trigonometric expression: $\cos {{25}^{\circ }}-\cos {{65}^{\circ }}=$.

& A.\sqrt{2}\cos {{20}^{\circ }} \\\ & B.\sqrt{2}\sin {{20}^{\circ }} \\\ & C.\sqrt{3}\cos {{20}^{\circ }} \\\ & D.\sqrt{3}\sin {{20}^{\circ }} \\\ \end{aligned}$$

Answer 1

In this question, we need to find the value of a function given in cosine angle. Since function is given in the form of $\cos C- \cos D$, so we will use the formula of subtraction of two cosine function given by: $\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)$. This will give us a function in the form of the multiplication of two sine functions. We will put the value of one of the sine functions known from the trigonometric ratio table and find our final answer.

Complete step-by-step solution
Here we are given the expression as $\cos {{25}^{\circ }}-\cos {{65}^{\circ }}$.
As we can see, the given expression is in the form of subtraction of two cosine function, so we can use the formula of subtraction of two cosine function given by: $\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)$.
Here let us take $C={{25}^{\circ }}\text{ and }D={{65}^{\circ }}$ so by formula we get:
$\cos {{25}^{\circ }}-\cos {{65}^{\circ }}=2\sin \left( \dfrac{{{25}^{\circ }}+{{65}^{\circ }}}{2} \right)\sin \left( \dfrac{{{65}^{\circ }}-{{25}^{\circ }}}{2} \right)$
${{25}^{\circ }}+{{65}^{\circ }}$ becomes equal to ${{90}^{\circ }}$ and dividing it by 2 gives us ${{45}^{\circ }}$. ${{65}^{\circ }}-{{25}^{\circ }}$ becomes equal to ${{40}^{\circ }}$ and dividing it by 2 gives us ${{20}^{\circ }}$. So we get:
$\begin{aligned} & \cos {{25}^{\circ }}-\cos {{65}^{\circ }}=2\sin \left( \dfrac{{{90}^{\circ }}}{2} \right)\sin \left( \dfrac{{{40}^{\circ }}}{2} \right) \\\ & \Rightarrow \cos {{25}^{\circ }}-\cos {{65}^{\circ }}=2\sin {{45}^{\circ }}\sin {{20}^{\circ }} \\\ \end{aligned}$
As we can see, none of our options match with the current answer, so let us try to simplify it now.
We know, in the trigonometric ratio table, we have values of angles ${{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}$. So we can have a value of $\sin {{45}^{\circ }}$.
We know value of $\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ so we get:
$\Rightarrow \cos {{25}^{\circ }}-\cos {{65}^{\circ }}=2\times \dfrac{1}{\sqrt{2}}\times \sin {{20}^{\circ }}$.
Since 2 can be written as $\sqrt{2}\times \sqrt{2}$ so we get:

& \Rightarrow \cos {{25}^{\circ }}-\cos {{65}^{\circ }}=\dfrac{\sqrt{2}\times \sqrt{2}}{\sqrt{2}}\times \sin {{20}^{\circ }} \\\ & \Rightarrow \cos {{25}^{\circ }}-\cos {{65}^{\circ }}=\sqrt{2}\sin {{20}^{\circ }} \\\ \end{aligned}$$ **Hence option B is the correct answer.** **Note:** Students should note that, in the formula, angle of sine is $\left( \dfrac{D-C}{2} \right)$ changing position of C and D will give us change in answer because $\sin \left( -\theta \right)=-\sin \theta $. Take care of signs while applying formulas of addition or subtraction of two cosine or two sine functions. Keep in mind all the values from the trigonometric ratio table.