Is it possible to get a formula that has a fixed number of terms for this ecuation:
$$\cos 3+\cos 7+...+\cos(2n+3)=?$$
For example: a product of trigonometric functions
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yes it is possible – Dr. Sonnhard Graubner Jan 20 '17 at 16:34
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Whats the formula then? – robertooverflow97 Jan 20 '17 at 16:35
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$\cos 3+\cos 7+...+\cos(2n+3)=?$ is false. – Nosrati Jan 20 '17 at 16:40
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2As it stands, your equation is impossible to interpret. Do you mean $\cos 3 + \cos 5+\dcots$? Or do you mean $\cos(4n+3)$? – TonyK Jan 20 '17 at 16:44
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One could joke that the formula is unbounded thus cannot be compact :D – N. S. Jan 20 '17 at 19:58
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Try Euler's formula. Miracles can happen. You will encounter a geometric series. But it only works for summation.
apprenant
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A direct reference: https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Other_sums_of_trigonometric_functions – apprenant Jan 20 '17 at 16:38
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$$S= \cos 3+\cos 7+...+\cos(2n+3) \\ 2\sin(2) S= 2\sin 2\left(\cos 3+\cos 7+...+\cos(2n+3) \right) \\ =2\sin2 \cos3 +2\sin2 \cos 7 + ... +2\sin2 \cos (2n+3)$$
Now, using the identity $$2\sin x \cos y = \sin(y+x)-\sin(y-x)$$ we get $$2\sin(2) S=(\sin 5-\sin1) + (\sin9-\sin 5) + ... +(\sin(2n+5)-\sin(2n+1) \\ 2\sin(2) S=\sin(2n+5)-\sin(1) $$
N. S.
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$${\frac {\cos \left( 5+2\,n \right) -\cos \left( 3+2\,n \right) -\cos \left( 3 \right) +\cos \left( 1 \right) }{2\,\cos \left( 2 \right)-2 }} $$
Robert Israel
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2Just taking a formula out of your sleeve is not that useful for the OP. Deriving the formula would be useful. – Wojowu Jan 20 '17 at 16:50
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It can be obtained using Euler's formula and a geometric progression, as apprenant noted. – Robert Israel Jan 20 '17 at 20:59