Đáp án:
$\begin{array}{l}
4)\\
\sin {30^0} + \sin {60^0} = \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2} = \dfrac{{1 + \sqrt 3 }}{2}\\
\cos {60^0} + \cos {30^0} = \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2} = \dfrac{{1 + \sqrt 3 }}{2}\\
\sin {60^0} + \cos {30^0} = \dfrac{{\sqrt 3 }}{2} + \dfrac{{\sqrt 3 }}{2} = \sqrt 3 \\
\sin {30^0} + \sin {60^0} + \cos {45^0}\\
= \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2} + \dfrac{{\sqrt 2 }}{2}\\
= \dfrac{{1 + \sqrt 3 + \sqrt 2 }}{2}\\
5)\\
a)\sin {42^0} = \cos \left( {{{90}^0} - {{42}^0}} \right)\\
= \cos {48^0} = \sqrt {1 - {{\sin }^2}{{48}^0}} \\
= \sqrt {1 - 0,{{7431}^2}} \\
= 0,669\\
b)\cos {46^0}\\
= \sin \left( {{{90}^0} - {{46}^0}} \right)\\
= \sin {44^0}\\
= \sqrt {1 - {{\cos }^2}{{44}^0}} \\
= \sqrt {1 - 0,{{7193}^2}} \\
= 0,695
\end{array}$
