Finding value of $\displaystyle \frac{\sum^{18}_{k=0}\sin^6(10k^\circ)}{\sum^{18}_{k=0}\cos^6(10k^\circ)}$
what i try
$\displaystyle \sin^6(x)=(\sin^3 x)^2=\frac{1}{16}\bigg[3\sin(x)-\sin(3x)\bigg]^2=\frac{1}{16}\bigg[9\sin^2(x)+\sin^2(3x)-6\sin (x)\sin(3x)\bigg]$
$\displaystyle \sin^6(x)=\frac{1}{16}\bigg[\frac{9}{2}-\frac{9}{2}\cos(2x)+\frac{1}{2}-\frac{1}{2}\cos(6x)-3\cos(2x)+3\cos(4x)\bigg]$
$\displaystyle \sin^6(x)=(\sin^3 x)^2=\frac{1}{16}\bigg[3\cos(x)+\cos(3x)\bigg]^2=\frac{1}{16}\bigg[9\cos^2(x)+\cos^2(3x)+6\cos(x)\cos(3x)\bigg]$
and $\displaystyle \cos^6(x)=\frac{1}{16}\bigg[\frac{9}{2}+\frac{9}{2}\cos(2x)+\frac{1}{2}+\frac{1}{2}\cos(6x)+3\cos(2x)+3\cos(4x)\bigg]$
How do i solve it help me please
The following identities can be used to evaluate the ratio $$\sin^6{(x)}=-\frac1{32}\cos{(6x)}+\frac3{16}\cos{(4x)}-\frac{15}{32}\cos{(2x)}+\frac5{16}$$ $$\cos^6{(x)}=\frac1{32}\cos{(6x)}+\frac3{16}\cos{(4x)}+\frac{15}{32}\cos{(2x)}+\frac5{16}$$ These can be found by expanding both $$(e^{ix}+e^{-ix})^6=(2\cos{(x)})^6$$ $$(e^{ix}-e^{-ix})^6=(2i\sin{(x)})^6$$ then applying De Moivre's theorem.
So the numerator summation becomes $$\sum_{k=0}^{18} \bigg(-\frac1{32}\cos{(\frac{k\pi}{3})}+\frac3{16}\cos{(\frac{2k\pi}{9})}-\frac{15}{32}\cos{(\frac{k\pi}{9})}+\frac5{16}\bigg)$$ As $$\sum_{k=0}^{18} \cos{(\frac{k\pi}3)}=\sum_{k=0}^{18} \cos{(\frac{2k\pi}9)}=\sum_{k=0}^{18} \cos{(\frac{k\pi}9)}=1$$ The summation simplifies to $$-\frac1{32}+\frac3{16}-\frac{15}{32}+\frac5{16}(19)=\frac{45}8$$
The denominator summation becomes $$\sum_{k=0}^{18} \bigg(\frac1{32}\cos{(\frac{k\pi}{3})}+\frac3{16}\cos{(\frac{2k\pi}{9})}+\frac{15}{32}\cos{(\frac{k\pi}{9})}+\frac5{16}\bigg)$$ which similarly simplifies to $$\frac1{32}+\frac3{16}+\frac{15}{32}+\frac5{16}(19)=\frac{53}8$$
So the ratio is then $$\frac{\big(\frac{45}8\big)}{\big(\frac{53}8\big)}=\frac{45}{53}$$