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Dимасuk
@Dимасuk
July 2022
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Зная, что [tex]sin18^{\circ} = \dfrac{ \sqrt{5}-1} {4} [/tex], найдите [tex]sin3^{\circ}[/tex]
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sedinalana
Sin18=(√5-1)/4
cos18=√(1-sin²18)=√(1-(6-2√5)/16)=√(10+2√5)/4
sin(45-30)=sin45cos30-cos45sin30=√2/2*√3/2-√2/2*1/2=(√6-√2)/4
cos(45-30)=cos45cos30+sin45sin30=√2/2*√3/2+√2/2*1/2=(√6+√2)/4
sin3=sin(18-15)=sin18cos15-cos18sin15=
=sin18*cos(45-30)-cos18*sin(45-30)=
=(√5-1)/4*(√6+√2)/4-√(10+2√5)/4*(√6-√2)/4=
=(√5-1)*(√6+√2)/16-√(10+2√5)*(√6-√2)/16
2 votes
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Dимасuk
Спасибо.
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Answers & Comments
cos18=√(1-sin²18)=√(1-(6-2√5)/16)=√(10+2√5)/4
sin(45-30)=sin45cos30-cos45sin30=√2/2*√3/2-√2/2*1/2=(√6-√2)/4
cos(45-30)=cos45cos30+sin45sin30=√2/2*√3/2+√2/2*1/2=(√6+√2)/4
sin3=sin(18-15)=sin18cos15-cos18sin15=
=sin18*cos(45-30)-cos18*sin(45-30)=
=(√5-1)/4*(√6+√2)/4-√(10+2√5)/4*(√6-√2)/4=
=(√5-1)*(√6+√2)/16-√(10+2√5)*(√6-√2)/16