Question

Выразите cos(70+a) через b если b=sin(40+a)

Answer 1

$b=\sin (40\textdegree+\alpha)$

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

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

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Использованы формулы

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