Question

Help please!

Answer 1

Answer:  see proof below

Step-by-step explanation:

Use the following Sum to Product Identities:

$\cos A+\cos B=2\cos \bigg(\dfrac{A+B}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B}{2}\bigg)\\\\\\\sin A-\sin B=2\cos \bigg(\dfrac{A+B}{2}\bigg)\cdot \sin \bigg(\dfrac{A-B}{2}\bigg)$

Use Even/Odd Identities:    cos (-A) = - cos A

                                             sin (-A) =   sin (A)

Use Quotient Identity:   $\cot A=\dfrac{\cos A}{\sin A}$

Proof LHS →  RHS:

$\text{LHS:}\qquad \qquad \qquad \dfrac{\cos x+\cos 7x}{\sin x - \sin 7x}$

$\text{Sum to Product:}\qquad \dfrac{2\cos \bigg(\dfrac{x+7x}{2}\bigg)\cdot \cos \bigg(\dfrac{x-7x}{2}\bigg)}{2\cos \bigg(\dfrac{x+7x}{2}\bigg)\cdot \sin \bigg(\dfrac{x-7x}{2}\bigg)}$

$\text{Simplify:}\qquad \qquad \quad \dfrac{2\cos \bigg(\dfrac{8x}{2}\bigg)\cdot \cos \bigg(\dfrac{-6x}{2}\bigg)}{2\cos \bigg(\dfrac{8x}{2}\bigg)\cdot \sin \bigg(\dfrac{-6x}{2}\bigg)}$

                          $= \dfrac{ \cos (-3x)}{\sin(-3x)}$

$\text{Even Odd:}\qquad \quad \dfrac{ -\cos (3x)}{\sin(3x)}$

$\text{Quotient:}\qquad \quad -\cot(3x)$

LHS = RHS  $\checkmark$