Trigonometribuktikan jika hasil dari [tex] \frac{ \sin( \alpha ) }{1 + \cos( \alpha ) } + \frac{ \si.... Question from @Zdytx

$\displaystyle \rm\frac{ \sin( \alpha ) }{1 + \cos( \alpha ) } + \frac{ \sin( \alpha ) }{ 1 - \cos( \alpha ) } = 2 \csc( \alpha )$

$\small \displaystyle \rm \left(\frac{ \sin( \alpha ) }{1 + \cos( \alpha ) } \times \frac{1 - \cos( \alpha ) }{1 - \cos( \alpha )} \right) + \left(\frac{ \sin( \alpha ) }{ 1 - \cos( \alpha ) } \times \frac{1 + \cos( \alpha ) }{1 + \cos( \alpha )} \right)= 2 \csc( \alpha )$

$\small\displaystyle\rm\frac{ \sin( \alpha ) - \sin( \alpha ) \cos( \alpha ) }{1 - \cos^{2} ( \alpha ) } + \frac{ \sin( \alpha )+ \sin( \alpha ) \cos( \alpha )}{ 1 - \cos^{2} ( \alpha ) } = 2 \csc( \alpha )$

$\small\displaystyle\rm\frac{ \sin( \alpha ) - \cancel{\sin( \alpha ) \cos( \alpha ) } }{1 - \cos^{2} ( \alpha ) } + \frac{ \sin( \alpha )+ \cancel{\sin( \alpha ) \cos( \alpha )}}{ 1 - \cos^{2} ( \alpha ) } = 2 \csc( \alpha )$

$\small\displaystyle\rm\frac{ \sin( \alpha ) }{1 - \cos^{2} ( \alpha ) } + \frac{ \sin( \alpha )}{ 1 - \cos^{2} ( \alpha ) } = 2 \csc( \alpha )$

$\small\displaystyle\rm\frac{ 2\sin( \alpha ) }{1 - \cos^{2} ( \alpha ) } = 2 \csc( \alpha )$

$\small\displaystyle\rm\frac{ 2\sin( \alpha ) }{ \sin^{2} ( \alpha ) } = 2 \csc( \alpha )$

$\small\displaystyle\rm\frac{ 2\sin( \alpha ) }{ \sin ( \alpha ) \sin( \alpha ) } = 2 \csc( \alpha )$

$\small\displaystyle\rm\frac{ 2 \cancel{\sin( \alpha ) }}{ \sin ( \alpha ) \cancel{\sin( \alpha ) } }= 2 \csc( \alpha )$

$\small\displaystyle\rm\frac{ 2 }{ \sin ( \alpha ) }= 2 \csc( \alpha )$

$\small\displaystyle\rm\frac{ 2 }{ \frac{1}{ \csc( \alpha ) } }= 2 \csc( \alpha )$

$\boxed {\small\displaystyle\rm2 \csc( \alpha ) = 2 \csc( \alpha ) } \: \: \: \rightarrow \: \: \: terbukti$

4 votes Thanks 5

Rixxel17 soalnya menurutku sama aja

Zdytx sorry² dh benar tadi baca nya ²/1/sin(α)

Rixxel17 oke kak

Zdytx saran, supaya gk terlalu panjang penyelesaiannya, sebaiknya baris ke-3, ke-5, & ke-8 dihilangkan aja

Rixxel17 iyaa kk, tpi aku suka yg panjang

Zdytx lha kasian kalo ada yg nanya soal gini, isi jwbannya terlalu panjang

Rixxel17 iya juga sih, jdi gk dapat like nanti. Org lebih pilih yg pendek :v

Zdytx klau mau di cancel terus dijumlahkan, lbih baik langsung cancel dan jumlahkan nya bersamaan

Zdytx btw keren jawabannya:)

Rixxel17 mksih kakk :)

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2 votes Thanks 4

SaputriCanz Makasih bestie :D

Zdytx bestie?

SaputriCanz yoi semua aku panggil bestie :)

SaputriCanz Like ppkn ku dong mba :v

SaputriCanz makasih mba -,-

SaputriCanz Mau Quiz ga)

SaputriCanz ??

InWorldSystema Diakun ini

Zdytx tdk perlu bestie

Zdytx oh ya, klo ketemu bentuk yg mirip (di penyebut [(1 + cos α)(1 - cos α)]) bisa lebih mudah pakai sifat aljabar [(a+b)(a-b) = a² - b²] --> [(1 + cos α)(1 - cos α) = 1 - cos² α = sin² α] jadi tdk perlu di ubah ke negatif

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