Croatian Mathematical Society Competitions

By applying the cosine theorem we get $$\cos \alpha =\frac{|CA{|}^2+|AB{|}^2-|BC{|}^2}{2|CA|\cdot |AB|}=\frac{25+16-49}{2\cdot 5\cdot 4}=-\frac{1}{5},$$ and therefore $$ $$\begin{array}{rl}{\sin }^6\frac{\alpha }{2}+{\cos }^6\frac{\alpha }{2}& =\left({\sin }^2\frac{\alpha }{2}+{\cos }^2\frac{\alpha }{2}\right)\left({\sin }^4\frac{\alpha }{2}-{\sin }^2\frac{\alpha }{2}{\cos }^2\frac{\alpha }{2}+{\cos }^4\frac{\alpha }{2}\right)\\ & ={\left({\sin }^2\frac{\alpha }{2}+{\cos }^2\frac{\alpha }{2}\right)}^2-3{\sin }^2\frac{\alpha }{2}{\cos }^2\frac{\alpha }{2}\\ & =1-\frac{3}{4}{\sin }^2\alpha \\ & =\frac{1}{4}+\frac{3}{4}{\cos }^2\alpha \\ & =\frac{1}{4}+\frac{3}{4}\cdot \frac{1}{25}=\frac{7}{25}.\end{array}$$