![sum(i=1)^(2n) sin^(-1)(xi)=npi then the value of sum(i=1)^n cos^(-1 )xi+sum(i=1)^n tan^(-1)xi= (A) (npi)/4 (B) (2/3)npi (C) (5/4)npi (D) 2npi sum(i=1)^(2n) sin^(-1)(xi)=npi then the value of sum(i=1)^n cos^(-1 )xi+sum(i=1)^n tan^(-1)xi= (A) (npi)/4 (B) (2/3)npi (C) (5/4)npi (D) 2npi](https://d10lpgp6xz60nq.cloudfront.net/ss/web/75564.jpg)
sum(i=1)^(2n) sin^(-1)(xi)=npi then the value of sum(i=1)^n cos^(-1 )xi+sum(i=1)^n tan^(-1)xi= (A) (npi)/4 (B) (2/3)npi (C) (5/4)npi (D) 2npi
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![functional analysis - Show that the exponentials $1, e^{2\pi ix}, \dots , e^{2\pi ikx}, \dots$ form the basis for trigonometric polynomials. - Mathematics Stack Exchange functional analysis - Show that the exponentials $1, e^{2\pi ix}, \dots , e^{2\pi ikx}, \dots$ form the basis for trigonometric polynomials. - Mathematics Stack Exchange](https://i.stack.imgur.com/tXPHW.png)
functional analysis - Show that the exponentials $1, e^{2\pi ix}, \dots , e^{2\pi ikx}, \dots$ form the basis for trigonometric polynomials. - Mathematics Stack Exchange
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![Statement-1: sin`(pi+theta)=-sin theta` Statement-2: `sin(npi+theta)=(-1)^(n) sin theta`, `n in N`. - YouTube Statement-1: sin`(pi+theta)=-sin theta` Statement-2: `sin(npi+theta)=(-1)^(n) sin theta`, `n in N`. - YouTube](https://i.ytimg.com/vi/-I9CGkRl5kk/maxresdefault.jpg)