Tentukan tan a + tan b

Umi Finch-Fletchley Alfiani

mohon bantuannya ..🙂
diket: \dfrac{\cos (a+b)}{\cos (a-b)} = \dfrac{3}{4} dan a+b =45^{\circ},
tentukan nilai \tan a + \tan b.

 

Rahman Setiawan

kalo gak salah itung hslnya
\frac{6}{7}. btl gk y?

 

Umi Finch-Fletchley Alfiani

hmm, caranya?🙂

 

Rahman Setiawan

\begin{array}{rl}a+b &= 45^{\circ}\\&\\  \cos (a+b) &= \dfrac{\sqrt{2}}{2}\\&\\  \cos a\cos b - \sin a\sin b &= \dfrac{\sqrt{2}}{2}\;\cdots\;(1)\end{array}

\begin{array}{rl}\dfrac{cos (a+b)}{cos (a-b)} &= \dfrac{3}{4}\\&\\  \cos (a-b) &= \dfrac{\cos (a+b)}{\frac{3}{4}}\\&\\  \cos (a-b) &= \dfrac{2\sqrt{2}}{3}\\&\\  \cos a \cos b + \sin a \sin b &= \dfrac{2\sqrt{2}}{3}\;\cdots\;(2)\end{array}

dr (1) dan (2):

\begin{array}{rl}\cos a\cos b &= \dfrac{7\sqrt{2}}{12}\\&\\  \sin a\sin b &= \dfrac{\sqrt{2}}{12}\\&\\  \tan(a+b) &= \dfrac{\tan a + \tan b}{1- \tan a\cdot\tan b}\\&\\  \tan a + \tan b &= \tan(a+b)\cdot(1- \tan a\cdot\tan b)\\&\\  \tan a + \tan b &= \tan45^{\circ}\left (1- \dfrac{\frac{\sqrt{2}}{12}}{\frac{7\sqrt{2}}{12}}\right )\\&\\  \tan a + \tan b &= 1\left (1- \dfrac{1}{7}\right ) = \dfrac{6}{7}.\end{array}

mhn koreksinya, kali ja sy kurang tliti.

 

Tulisan Terbaru :

Tentang msihabudin

Just a crazy people
Pos ini dipublikasikan di SOUL-MATE-MATIKA dan tag . Tandai permalink.

Tinggalkan Balasan

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout / Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout / Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout / Ubah )

Foto Google+

You are commenting using your Google+ account. Logout / Ubah )

Connecting to %s