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Question: Prove that, \(\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2...

Prove 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.

Explanation

Solution

Hint: We will first start by using the identity that cos⁡C−cos⁡D=2sin⁡(C+D2)sin⁡(D−C2)\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right). Then we will use the fact that sin⁡(π−θ)=sin⁡θ\sin \left( \pi -\theta \right)=\sin \theta to find the value of sin⁡(3π4)\sin \left( \dfrac{3\pi }{4} \right). Then we will use this value to prove the left hand side of the equation to be equal to the right side.

Complete step-by-step answer:
Now, we have been given to prove 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.
Now, we will take the left hand side of the equation and prove it to be equal to the right hand side. So, in left hand side we have,
cos⁡(3π4+x)−cos⁡(3π4−x)\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)
Now, we know that the trigonometric identity that,
cos⁡C−cos⁡D=2sin⁡(C+D2)sin⁡(D−C2)\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)
Now, we have the left hand side as,
⇒cos⁡(3π4+x)−cos⁡(3π4−x)=2sin⁡(6π8)sin⁡(−2x2) =2sin⁡(3π4)sin⁡(−x) \begin{aligned} & \Rightarrow \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=2\sin \left( \dfrac{6\pi }{8} \right)\sin \left( \dfrac{-2x}{2} \right) \\\ & =2\sin \left( \dfrac{3\pi }{4} \right)\sin \left( -x \right) \\\ \end{aligned}
Now, we know the trigonometric identity that,
sin⁡(−x)=−sin⁡(x) sin⁡(π−θ)=sin⁡(θ) \begin{aligned} & \sin \left( -x \right)=-\sin \left( x \right) \\\ & \sin \left( \pi -\theta \right)=\sin \left( \theta \right) \\\ \end{aligned}
So, using this we have,
⇒cos⁡(3π4+x)−cos⁡(3π4−x)=2sin⁡(π−π4)(−sin⁡x) =−2sin⁡(π4)sin⁡(x) \begin{aligned} & \Rightarrow \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=2\sin \left( \pi -\dfrac{\pi }{4} \right)\left( -\sin x \right) \\\ & =-2\sin \left( \dfrac{\pi }{4} \right)\sin \left( x \right) \\\ \end{aligned}
Now, we know the fact that sin⁡(π4)=12\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}.So, we have,
⇒cos⁡(3π4+x)−cos⁡(3π4−x)=−2×12sin⁡x =−2sin⁡x \begin{aligned} & \Rightarrow \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\times \dfrac{1}{\sqrt{2}}\sin x \\\ & =-\sqrt{2}\sin x \\\ \end{aligned}
Hence, we have LHS = RHS and therefore, it is 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: It is important to note that to find the value of sin⁡(3π4)\sin \left( \dfrac{3\pi }{4} \right) we have re-written it as sin⁡(4π−π4) or sin⁡(π−π4)\sin \left( \dfrac{4\pi -\pi }{4} \right)\ or\ \sin \left( \pi -\dfrac{\pi }{4} \right) and then used the fact that sin⁡(π−θ)=sin⁡θ\sin \left( \pi -\theta \right)=\sin \theta to find its value. Also, we have used the property of sine that sin⁡(−x)=−sin⁡x\sin \left( -x \right)=-\sin x to finalize the proof.