if 8x=π then show that cos 7x + cos x. = 0
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Answered by
84
Here is your answer.....
Given, 8x = π
⇒ x = π/8
Now, taking LHS
cos 7x + cos x
= cos 7π/8 + cos π/8
= cos (π - π/8) + cos π/8
= - cos π/8 + cos π/8 [As, cos (π - π/8) = -cos π/8 ]
= 0 = RHS
Hence, shown
I hope it helps you ^_^
Given, 8x = π
⇒ x = π/8
Now, taking LHS
cos 7x + cos x
= cos 7π/8 + cos π/8
= cos (π - π/8) + cos π/8
= - cos π/8 + cos π/8 [As, cos (π - π/8) = -cos π/8 ]
= 0 = RHS
Hence, shown
I hope it helps you ^_^
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