Forum Sains Indonesia

Pendidikan dan Karir => Bimbingan Belajar => Bimbel Matematika => Topik dimulai oleh: CrisTaNa pada Agustus 25, 2010, 08:03:29 PM

Judul: 3gonome3
Ditulis oleh: CrisTaNa pada Agustus 25, 2010, 08:03:29 PM
jika \frac{\cos \alpha}{1-\sin \alpha}=a, nyatakan \tan \frac12 \alpha dalam a
Judul: Re: 3gonome3
Ditulis oleh: Mtk Kerajaan Mataram pada Agustus 26, 2010, 01:19:17 AM
\frac{\cos \alpha}{1-\sin \alpha}=a

==> \frac{\(\cos \frac{\alpha}{2}\)^2-\(\sin \frac{\alpha}{2}\)^2}{\(\cos \frac{\alpha}{2}\)^2+\(\sin \frac{\alpha}{2}\)^2-2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}=a

==> \frac{\(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\)\(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\)}{\(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\)\(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\)}=a

==> \frac{\(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\)}{\(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\)}=a

==> \frac{\frac{\(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\)}{\cos \frac{\alpha}{2}}}{\frac{\(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\)}{\cos \frac{\alpha}{2}}}=a

==> \frac{\(1-\tan \frac{\alpha}{2}\)}{\(1+\tan \frac{\alpha}{2}\)}=a

==> 1-\tan \frac{\alpha}{2}=a+a\tan \frac{\alpha}{2}

==> \tan \frac{\alpha}{2}=\frac{1-a}{1+a}
Judul: Re: 3gonome3
Ditulis oleh: PocongSains pada September 04, 2010, 12:09:23 AM
a = 0

tan(1/2)a = a