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Question: Prove the trigonometric function: \(\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi...

Prove the trigonometric function:
cos⁡(3π4+x)−cos⁡(3π4−x)=−2sin⁡(x)\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin (x)

Explanation

Solution

Hint:In this question, we have to find the sum of cosines of two different angles. However, the angles in each term are a sum of two angles. Therefore, we should understand the formula for finding the cosine of the sum of two angles and then use it to solve this question.

Complete step-by-step answer:
Here, we are given the cosines of two different angles and we have to find their sum. However, the angle in each cosine term is itself a sum or difference of two angles. Therefore, we should first understand the equation for finding the cosine of sum and difference of two angles.
We know that the expansion of cosine of sum and difference of two angles is given by
cos⁡(a+b)=cos⁡(a)cos⁡(b)−sin⁡(a)sin⁡(b)........(1.1) cos⁡(a−b)=cos⁡(a)cos⁡(b)+sin⁡(a)sin⁡(b)............(1.2) \begin{aligned} & \cos (a+b)=\cos (a)\cos (b)-\sin (a)\sin (b)........(1.1) \\\ & \cos (a-b)=\cos (a)\cos (b)+\sin (a)\sin (b)............(1.2) \\\ \end{aligned}

If we subtract these two equations, we obtain
cos⁡(a+b)−cos⁡(a−b)=cos⁡(a)cos⁡(b)−sin⁡(a)sin⁡(b)−(cos⁡(a)cos⁡(b)+sin⁡(a)sin⁡(b)) ⇒cos⁡(a+b)−cos⁡(a−b)=−2sin⁡(a)sin⁡(b)....................(1.3) \begin{aligned} & \cos (a+b)-\cos (a-b)=\cos (a)\cos (b)-\sin (a)\sin (b)-\left( \cos (a)\cos (b)+\sin (a)\sin (b) \right) \\\ & \Rightarrow \cos (a+b)-\cos (a-b)=-2\sin (a)\sin (b)....................(1.3) \\\ \end{aligned}

We find that this equation is similar to the expression in the question. Taking a=3π4a=\dfrac{3\pi }{4}and b=xb=x in equation (1.3), we obtain
cos⁡(3π4+x)−cos⁡(3π4−x)=−2sin⁡(3π4)sin⁡(x).......(1.4)\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\sin \left( \dfrac{3\pi }{4} \right)\sin (x).......(1.4)
Now, using the trigonometric equation
sin⁡(π2+x)=sin⁡(x)\sin \left( \dfrac{\pi }{2}+x \right)=\sin (x)
We obtain
sin⁡(3π4)=sin⁡(π2+π4)=sin⁡(π4)=12.........(1.5)\sin \left( \dfrac{3\pi }{4} \right)=\sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}.........(1.5)
Thus, using this value in equation (1.4), we obtain
cos⁡(3π4+x)−cos⁡(3π4−x)=−2×12×sin⁡(x)=−2sin⁡(x)...........(1.6)\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\times \dfrac{1}{\sqrt{2}}\times \sin (x)=-\sqrt{2}\sin (x)...........(1.6)
The right hand side of equation (1.6) is the same as the right hand side of the question. Hence, we have proved that
cos⁡(3π4+x)−cos⁡(3π4−x)=−2sin⁡(x)\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin (x)

Note: In this question, we did not expand the cosine terms separately but instead used equation (1.3) to evaluate the question. One can expand the cosine terms separately. However, then two terms involving the cosine terms will get cancelled out to give the same answer as we have obtained.