Монгол Бодлогын Сан

Заавар:

Бодолт: $$\begin{align*} \sin^4\alpha-\cos^4\alpha&=(\sin^2\alpha-\cos^2\alpha)(\sin^2\alpha+\cos^2\alpha)\\ &=\sin^2\alpha-\cos^2\alpha\\ &=\left(\dfrac{2\tg\dfrac{\alpha}{2}}{1+\tg^2\dfrac{\alpha}{2}}\right)^2-\left(\dfrac{1-\tg^2\dfrac{\alpha}{2}}{1+\tg^2\dfrac{\alpha}{2}}\right)^2\\ &=\left(\dfrac{2\cdot\frac12}{1+\left(\frac{1}{2}\right)^2}\right)^2-\left(\dfrac{1-\left(\frac{1}{2}\right)^2}{1+\left(\frac{1}{2}\right)^2}\right)^2\\ &=\left(\dfrac{1}{5/4}\right)^2-\left(\dfrac{3/4}{5/4}\right)^2=\dfrac{16}{25}-\dfrac{9}{25}=0.28 \end{align*}$$