выразите cos(70+a) через b если b=sin(40+a) — Правильный ответ на вопрос найдете ниже

Ответ:

[tex]b=\sin (40\textdegree+\alpha)[/tex]

[tex]\cos (70\textdegree+\alpha)=\cos \Big(30\textdegree+(40\textdegree+\alpha)\Big)=\\\\=\cos 30\textdegree\cdot \cos(40\textdegree+\alpha) -\sin30\textdegree\cdot \sin(40\textdegree+\alpha)=\\\\=\dfrac{\sqrt3}2\cdot \cos(40\textdegree+\alpha) -\dfrac 12\cdot \sin(40\textdegree+\alpha)=\\\\=\pm\dfrac{\sqrt3}2\cdot \sqrt{1-\sin^2(40\textdegree+\alpha)} -\dfrac 12\cdot \sin(40\textdegree+\alpha)=\\\\=\pm\dfrac{\sqrt3}2\cdot \sqrt{1-b^2} -\dfrac 12\cdot b[/tex]

[tex]\boldsymbol{\cos (70\textdegree+\alpha)=\dfrac 12\bigg(\pm\sqrt {3\Big(1-b^2\Big)}-b\bigg)}[/tex]

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Использованы формулы

[tex]\cos (\alpha +\beta )=\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta \\\\\cos ^2 \alpha +\sin^2=1~~~\Rightarrow~~~\cos\alpha =\pm\sqrt{1-\sin^2\alpha }[/tex]