Ответ:
Применяем формулу понижения степени [tex]\bf cos^2a=\dfrac{1+cos2a}{2}[/tex] .
[tex]\bf cos^215^\circ -cos^275^\circ =\dfrac{1+cos30^\circ }{2}-\dfrac{1+cos150^\circ }{2}=\\\\\\=\dfrac{1}{2}\cdot (1+cos30^\circ -1-cos(180^\circ -30^\circ \, )=\\\\\\=\dfrac{1}{2}\cdot (1+\dfrac{\sqrt3}{2}-1+cos30^\circ )=\dfrac{1}{2}\cdot \Big(\dfrac{\sqrt3}{2}+\dfrac{\sqrt3}{2}\Big)= \dfrac{\sqrt3}{2}[/tex]
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Ответ:
Применяем формулу понижения степени [tex]\bf cos^2a=\dfrac{1+cos2a}{2}[/tex] .
[tex]\bf cos^215^\circ -cos^275^\circ =\dfrac{1+cos30^\circ }{2}-\dfrac{1+cos150^\circ }{2}=\\\\\\=\dfrac{1}{2}\cdot (1+cos30^\circ -1-cos(180^\circ -30^\circ \, )=\\\\\\=\dfrac{1}{2}\cdot (1+\dfrac{\sqrt3}{2}-1+cos30^\circ )=\dfrac{1}{2}\cdot \Big(\dfrac{\sqrt3}{2}+\dfrac{\sqrt3}{2}\Big)= \dfrac{\sqrt3}{2}[/tex]