Question

Those are not the powers question from transformation of product and sum trigonometry

prove that : cos3A+cos5A+cos7A+cos15A= 4cos4Acos5Acos6A​

Answer 1

see explanation

Explanation:

Using the sum to product identity

cosx + cosy = 2cos(
`(x+y)/(2)` )cos(
`(x-y)/(2)` )

Consider left side

(cos5A +cos3A) + (cos15A + cos7A)

= 2cos(
`(5A+3A)/(2)` )cos(
`(5A-3A)/(2)` ) + 2cos(
`(15A+7A)/(2)`) cos(
`(15A-7A)/(2)` )

= 2cos(
`(8A)/(2)`) cos(
`(2A)/(2)`) + 2cos(
`(22A)/(2)` )cos(
`(8A)/(2)` )

= 2cos4AcosA + 2cos11A cos4A ← factor out 2cos4A from both terms

= 2cos4A( cos11A + cosA) ← repeat the process

= 2cos4A( 2cos(
`(11A+A)/(2)` )cos(
`(11A-A)/(2)` )

= 2cos4A(2cos(
`(12A)/(2)`)cos(
`(10A)/(2)` )

= 2cos4A(2cos6A cos5A)

= 4cos4Acos5Acos6A

= right side , thus verified