Linearise-cos-2-p-1-t- Tinku Tara June 4, 2023 Trigonometry 0 Comments FacebookTweetPin Question Number 103314 by Ar Brandon last updated on 14/Jul/20 Linearisecos2(p−1)(t) Answered by OlafThorendsen last updated on 14/Jul/20 cos2(p−1)(t)=[eit+e−it2]2(p−1)cos2(p−1)(t)=14p−1∑2(p−1)k=0C2(p−1)keikte−it(2p−2−k)cos2(p−1)(t)=14p−1∑2(p−1)k=0C2(p−1)ke−it(2p−2−2k)cos2(p−1)(t)=14p−1∑2(p−1)k=0C2(p−1)ke2it(k+1−p)cos2(p−1)(t)=(2p−2)!4p−1∑2(p−1)k=0cos[2(k+1−p)t]k!(2p−2−k)! Commented by Ar Brandon last updated on 14/Jul/20 Thanks Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Find-the-shortest-distance-between-the-curves-9x-2-9y-2-30y-16-0-and-y-2-x-3-Next Next post: Question-168855 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![cos^(2(p−1)) (t) = [((e^(it) +e^(−it) )/2)]^(2(p−1)) cos^(2(p−1)) (t) = (1/4^(p−1) )Σ_(k=0) ^(2(p−1)) C_(2(p−1)) ^k e^(ikt) e^(−it(2p−2−k)) cos^(2(p−1)) (t) = (1/4^(p−1) )Σ_(k=0) ^(2(p−1)) C_(2(p−1)) ^k e^(−it(2p−2−2k)) cos^(2(p−1)) (t) = (1/4^(p−1) )Σ_(k=0) ^(2(p−1)) C_(2(p−1)) ^k e^(2it(k+1−p)) cos^(2(p−1)) (t) = (((2p−2)!)/4^(p−1) )Σ_(k=0) ^(2(p−1)) ((cos[2(k+1−p)t])/(k!(2p−2−k)!))](https://www.tinkutara.com/question/Q103316.png)
