Question

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Answer 1

$1)\ \ 4sin^21^\circ \cdot cos^21^\circ -cos^22^\circ =(2sin1^\circ \cdot cos1^\circ )^2-cos^22^\circ =(sin2^\circ )^2-cos^22^\circ =\\\\=sin^22^\circ -cos^22^\circ =-(cos^22^\circ -sin^22^\circ )=-cos4^\circ \\\\\\2)\ \ (sin10^\circ +cos80^\circ )(cos80^\circ -cos10^\circ )=\\\\=sin10^\circ \cdot cos80^\circ -sin10^\circ \cdot cos10^\circ +cos^280^\circ -cos80^\circ \cdot cos10^\circ =$

$=\dfrac{1}{2}\cdot (sin90^\circ -sin70^\circ )-\dfrac{1}{2}\cdot sin20^\circ +\dfrac{1}{2}\cdot (1+cos160^\circ )-\dfrac{1}{2}\cdot (cos90^\circ +cos70^\circ )=\\\\=\dfrac{1}{2}\cdot (1-sin70^\circ -sin20^\circ +1+cos(180^\circ -20^\circ )-0-cos70^\circ )=\\\\=\dfrac{1}{2}\cdot (2-sin70^\circ -sin20^\circ -cos20^\circ -cos70^\circ )=\\\\=1-\dfrac{1}{2}\cdot (sin70^\circ +sin20^\circ )-\dfrac{1}{2}\cdot (cos70^\circ +cos20^\circ )=\\\\=1-sin45^\circ \cdot cos25^\circ -cos45^\circ \cdot cos25^\circ =$

$=1-\dfrac{\sqrt2}{2}\cdot cos25^\circ -\dfrac{\sqrt2}{2}\cdot cos25^\circ =1-\sqrt2\cdot cos25^\circ$