\displaystyle{\sin{{\left(\frac{\pi}{{24}}\right)}}}=\frac{{1}}{{2}}\sqrt{{{2}-\sqrt{{{2}+\sqrt{{3}}}}}} Explanation: As \displaystyle\frac{\pi}{{24}}=\frac{{180}^{\circ}}{{24}}={\left({7}\frac{{1}}{{2}}\right)}^{\circ} ...

The answer to this question depends on exactly what you mean by expressed in radicals. In the sense which is usually meant in Galois theory courses, \cos \pi/25 is expressible in radicals, but in a ...

In the sum of angle theorems, let a=b so that \cos(2a)=\cos^2(a)-\sin^2(a) By the last identity, notice that \cos^2(a)-\sin^2(a)=2\cos^2(a)-1 \cos^2(a)-\sin^2(a)=1-2\sin^2(a) Now let a=\pi/4 ...

By repeated application of angle sum formulas we may get, \sin (5x)=\sin^5 x+5 \cos^4 x\sin x-10 \sin^3 x \cos^2 x Let x=\frac{\pi}{5} and let \sin (\frac{\pi}{5})=u then we have, 0=u^5+5(1-u^2)^2 u-10(1-u^2)u^3 ...

Hint: from \cos(2(\frac{\pi}{3})+\frac{\pi}{3})= \cos(\pi)=-1, using summation and double-angle formulas we have: \left(2\cos^2(\pi/3)-1 \right)\cos(\pi/3)-2\left(1-\cos^2(\pi/3)\right)\cos(\pi/3)+1=0 ...

Note the pattern: \sin 0^{\circ} = \frac{\sqrt{0}}{2} \sin 30^{\circ} = \frac{\sqrt{1}}{2} \sin 45^{\circ} = \frac{\sqrt{2}}{2} \sin 60^{\circ} = \frac{\sqrt{3}}{2} \sin 90^{\circ} = \frac{\sqrt{4}}{2} ...