Question

Help please!

I need to prove this using identites
show all steps

cosx+cos7x/sinx-sin7x=-cot3x​

Answer 1

Answer: see proof below

Explanation:

Use the following Sum to Product Identities:

`\cos A+\cos B=2\cos \bigg((A+B)/(2)\bigg)\cdot \cos \bigg((A-B)/(2)\bigg)\\\\\\\sin A-\sin B=2\cos \bigg((A+B)/(2)\bigg)\cdot \sin \bigg((A-B)/(2)\bigg)`

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

sin (-A) = sin (A)

Use Quotient Identity:
`\cot A=(\cos A)/(\sin A)`

Proof LHS → RHS:

`\text{LHS:}\qquad \qquad \qquad (\cos x+\cos 7x)/(\sin x - \sin 7x)`

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

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

`= ( \cos (-3x))/(\sin(-3x))`

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

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

LHS = RHS
`\checkmark`