Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Using ${}^n{C}_k={}^{n-1}{C}_{k-1}+{}^{n-1}{C}_k,f(n)={\sum }_{k=0}^n\left({}^{n-1}{C}_{k-1}+{}^{n-1}{C}_k\right)\cos \frac{2k\pi }{n}=(g(n)-1)+(1+g(n))=2g(n)$ (a-t). $f(6)={\sum }_{k=0}^6{}^6{C}_k\cos \frac{k\pi }{3}=1+6\left(\frac{1}{2}\right)+15\left(-\frac{1}{2}\right)+20(-1)+15\left(-\frac{1}{2}\right)+6\left(\frac{1}{2}\right)+1=-27$, which is rational (b-p,r). $g(6)={\sum }_{k=0}^5{}^5{C}_k\cos \frac{k\pi }{3}$ involves only $\pm 1,\pm \frac{1}{2}$, hence rational (c-p). $g(8)={\sum }_{k=0}^7{}^7{C}_k\cos \frac{k\pi }{4}$ involves odd multiples of $\frac{\sqrt{2}}{2}$, hence irrational (d-q).