sitifatimahhubadillah Posted April 4, 2017 Report Share Posted April 4, 2017 Hi, anyone can help me how to prove this question using mathematical induction? i have been searching at some reference books at my university and i cant found the solutions. 1/2 + cosx +cos2x + ... + cosnx = [sin(n+1/2)x]/[2sin(x/2)] Reply Link to post Share on other sites More sharing options...
kw0573 Posted April 4, 2017 Report Share Posted April 4, 2017 (edited) This is the Lagrange's trig identity. I think the key to prove by induction is to convert sin (n+3/2)x into sin ((n+1) + 1/2) x = sin (n+1)x cos (x/2) + cos (n+1)x * sin (x/2), but to also do the same for sin (n+1/2)x as sin ((n+1) - 1/2)x Proof f(n)--> f(n+1) sin (n+1/2)x / (2 sin (x/2)) + cos (n+1)x ... use sin (a - b), where a = (n + 1)x, b = x/2 = [sin (n+1)x cos (x/2) - cos (n+1)x sin (x/2) + 2 cos(n+1)x sin (x/2)] / (2 sin (x/2)) = [sin (n+1)x cos(x/2) + cos(n+1)x sin (x/2)] / (2 sin (x/2)) = sin (n+3/2) x / (2 sin (x/2)) QED EDIT: If you are trying to prove f(1), then maybe tan (x / 2) = sin x / (1 + cos x) is helpful. Edited April 4, 2017 by kw0573 3 Reply Link to post Share on other sites More sharing options...
sitifatimahhubadillah Posted April 4, 2017 Author Report Share Posted April 4, 2017 Okayy . Thanks @kw0573 Reply Link to post Share on other sites More sharing options...
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