Answer:
see explanation
Step-by-step explanation:
Using the sum to product identity
cosx + cosy = 2cos([tex]\frac{x+y}{2}[/tex] )cos( [tex]\frac{x-y}{2}[/tex] )
Consider left side
(cos5A +cos3A) + (cos15A + cos7A)
= 2cos([tex]\frac{5A+3A}{2}[/tex] )cos([tex]\frac{5A-3A}{2}[/tex] ) + 2cos( [tex]\frac{15A+7A}{2}[/tex]) cos( [tex]\frac{15A-7A}{2}[/tex] )
= 2cos([tex]\frac{8A}{2}[/tex]) cos([tex]\frac{2A}{2}[/tex]) + 2cos([tex]\frac{22A}{2}[/tex] )cos([tex]\frac{8A}{2}[/tex] )
= 2cos4AcosA + 2cos11A cos4A ← factor out 2cos4A from both terms
= 2cos4A( cos11A + cosA) ← repeat the process
= 2cos4A( 2cos([tex]\frac{11A+A}{2}[/tex] )cos([tex]\frac{11A-A}{2}[/tex] )
= 2cos4A(2cos([tex]\frac{12A}{2}[/tex])cos([tex]\frac{10A}{2}[/tex] )
= 2cos4A(2cos6A cos5A)
= 4cos4Acos5Acos6A
= right side , thus verified
