# Trig

• Oct 21st 2013, 05:53 PM
srirahulan
Trig
Use only cos rule to prove this $\displaystyle cotA:cotB:cotC$=$\displaystyle (b^2+c^2-a^2)$:$\displaystyle (a^2+c^2-b^2)$:$\displaystyle (a^2+b^2-c^2)$

In this case i use cos rule but, after few step i cannot simplify, but i have finish this sum with sine rule, If anyone have a idea to do simplify this sum through cos rule>>>>>>
• Oct 22nd 2013, 01:00 AM
chiro
Re: Trig
Hey srirahulan.

Since cot A = cos(A)/sin(A), what are you using to calculate the sines if you can only use the cosine rule?
• Oct 22nd 2013, 08:04 AM
srirahulan
Re: Trig
i cannot understand, what you say?
• Oct 22nd 2013, 09:16 AM
SlipEternal
Re: Trig
Quote:

Originally Posted by srirahulan
Use only cos rule to prove this $\displaystyle cotA:cotB:cotC$=$\displaystyle (b^2+c^2-a^2)$:$\displaystyle (a^2+c^2-b^2)$:$\displaystyle (a^2+b^2-c^2)$

In this case i use cos rule but, after few step i cannot simplify, but i have finish this sum with sine rule, If anyone have a idea to do simplify this sum through cos rule>>>>>>

According to the Law of Cosines,
\displaystyle \begin{align*}a^2 & = b^2+c^2-2bc\cos A \\ \Rightarrow \cos A & = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{\mbox{adj}}{\mbox{hyp}} \\ \Rightarrow \cot A & = \dfrac{\mbox{adj}}{\mbox{opp}} = \dfrac{b^2+c^2-a^2}{\sqrt{4b^2c^2-(b^2+c^2-a^2)^2}} \\ b^2 & = a^2+c^2-2ac\cos B \\ \Rightarrow \cos B & = \dfrac{a^2+c^2-b^2}{2ac} = \dfrac{\mbox{adj}}{\mbox{hyp}} \\ \Rightarrow \cot B & = \dfrac{\mbox{adj}}{\mbox{opp}} = \dfrac{a^2+c^2-b^2}{\sqrt{4a^2c^2-(a^2+c^2-b^2)^2}} \\ c^2 & = a^2+b^2-2ab\cos C \\ \Rightarrow \cos C & = \dfrac{a^2+b^2-c^2}{2ab} = \dfrac{\mbox{adj}}{\mbox{hyp}} \\ \Rightarrow \cot C & = \dfrac{\mbox{adj}}{\mbox{opp}} = \dfrac{a^2+b^2-c^2}{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}} \end{align*}

Show that the denominators for $\displaystyle \cot A, \cot B, \cot C$ are all equal.
• Oct 22nd 2013, 04:52 PM
srirahulan
Re: Trig
Great Thanks For you>>>>