## Task A-3.1.

Given a triangle $A B C$ such that $|A B|=4,|B C|=7,|A C|=5$. Let $\alpha=\varangle B A C$. Calculate

$$
\sin ^{6} \frac{\alpha}{2}+\cos ^{6} \frac{\alpha}{2}
$$
## First Solution.

Notice that

$$
\begin{aligned}
\sin ^{6} \frac{\alpha}{2}+\cos ^{6} \frac{\alpha}{2} & =\left(\sin ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}\right)\left(\sin ^{4} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2}+\cos ^{4} \frac{\alpha}{2}\right) \\
& =\left(\sin ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}\right)^{2}-3 \sin ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2} \\
& =1-\frac{3}{4} \sin ^{2} \alpha \\
& =\frac{1}{4}+\frac{3}{4} \cos ^{2} \alpha
\end{aligned}
$$

By applying the cosine rule to triangle $ABC$, we get

$$
\cos \alpha=\frac{|A B|^{2}+|A C|^{2}-|B C|^{2}}{2|A C| \cdot|A B|}=\frac{16+25-49}{2 \cdot 5 \cdot 4}=-\frac{1}{5}
$$

Therefore, $\sin ^{6} \frac{\alpha}{2}+\cos ^{6} \frac{\alpha}{2}=\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{25}=\frac{7}{25}$.