Question

Please solve this question.​

Answer 1

Answer: see proof below

Explanation:

Use the Product to Sum Identity: cos A · cos B = [cos (A - B) + cos (A + B)]/2

Use the Odd Function Identity: cos (-A) = - cos (A)

Proof LHS → RHS:

`\text{LHS:}\qquad \qquad 2\cos \bigg((11\pi)/(16)\bigg)\cdot \cos \bigg((\pi)/(16)\bigg)+\cos \bigg((\pi)/(4)\bigg)+\cos \bigg((3\pi)/(8)\bigg)`

`\text{Prod to Sum:}\quad (2\bigg[\cos \bigg(((11\pi -\pi))/(16)\bigg)+\cos \bigg(((11\pi+\pi)/(16)\bigg)\bigg])/(2)+\cos \bigg((\pi)/(4)\bigg)+\cos \bigg((3\pi)/(8)\bigg)`

`=\cos \bigg((10\pi)/(16)\bigg)+\cos \bigg((12\pi)/(16)\bigg)+\cos \bigg((\pi)/(4)\bigg)+\cos \bigg((3\pi)/(8)\bigg)`

`=\cos \bigg((5\pi)/(8)\bigg)+\cos \bigg((3\pi)/(4)\bigg)+\cos \bigg((\pi)/(4)\bigg)+\cos \bigg((3\pi)/(8)\bigg)`

`=\cos \bigg((-3\pi)/(8)\bigg)+\cos \bigg((-\pi)/(4)\bigg)+\cos \bigg((\pi)/(4)\bigg)+\cos \bigg((3\pi)/(8)\bigg)`

`\text{Odd Functions:}\quad -\cos \bigg((3\pi)/(8)\bigg)-\cos \bigg((\pi)/(4)\bigg)+\cos \bigg((\pi)/(4)\bigg)+\cos \bigg((3\pi)/(8)\bigg)`

= 0

LHS = RHS
`\checkmark`