# Thread: Trigonometry - Compound Angles

1. ## Trigonometry - Compound Angles

Hi guys i've attempted this question 3 times now yet i keep getting the wrong answer so i turn to you my peers for help!.
Question:
Given that $cot(30°-\theta)=3$, find the value of $cot(\theta)$
My Workings Out:
Using $tan(30°-\theta)=\frac{tan(30°)-tan(\theta)}{1+tan(30°).tan(\theta)}$
$cot(30°-\theta)=\frac{1+tan(30°).tan(\theta)}{tan(30°)-tan(\theta)}$
$3=\frac{1+tan(30°).tan(\theta)}{tan(30°)-tan(\theta)}$
now i did tan30 which gives $\frac{\sqrt3}{3}$ i get
$3=\frac{1+\frac{\sqrt3}{3}tan(\theta)}{\frac{\sqrt 3}{3}-tan(\theta)}$
$3(\frac{\sqrt3}{3}-tan(\theta))=1+\frac{\sqrt3}{3}tan(\theta)$
$\sqrt3-3tan(\theta)=1+\frac{\sqrt3}{3}tan\theta$
re-arrange gives me
$-1+\sqrt3=\frac{9+\sqrt3}{3}tan(\theta)$
$tan\theta=\frac{-1+\sqrt3}{\frac{9+\sqrt3}{3}}=\frac{-29+9\sqrt3}{26}$
so
$cot\theta=\frac{26}{-26+9\sqrt3}$
$\frac{1+3\sqrt3}{3-\sqrt3}=\frac{6+5\sqrt3}{3}$

2. Originally Posted by Kevlar
Hi guys i've attempted this question 3 times now yet i keep getting the wrong answer so i turn to you my peers for help!.
Question:
Given that $cot(30°-\theta)=3$, find the value of $cot(\theta)$
My Workings Out:
Using $tan(30°-\theta)=\frac{tan(30°)-tan(\theta)}{1+tan(30°).tan(\theta)}$
$cot(30°-\theta)=\frac{1+tan(30°).tan(\theta)}{tan(30°)-tan(\theta)}$
$3=\frac{1+tan(30°).tan(\theta)}{tan(30°)-tan(\theta)}$
now i did tan30 which gives $\frac{\sqrt3}{3}$ i get
$3=\frac{1+\frac{\sqrt3}{3}tan(\theta)}{\frac{\sqrt 3}{3}-tan(\theta)}$
$3(\frac{\sqrt3}{3}-tan(\theta))=1+\frac{\sqrt3}{3}tan(\theta)$
$\sqrt3-3tan(\theta)=1+\frac{\sqrt3}{3}tan\theta$
re-arrange gives me
$-1+\sqrt3=\frac{9+\sqrt3}{3}tan(\theta)$
$tan\theta=\frac{-1+\sqrt3}{\frac{9+\sqrt3}{3}}=\frac{-29+9\sqrt3}{26}$
so
$cot\theta=\frac{26}{-26+9\sqrt3}$
$\frac{1+3\sqrt3}{3-\sqrt3}=\frac{6+5\sqrt3}{3}$
That was a lot of work! The first thing I saw was that this is wrong:

$tan\theta=\frac{-1+\sqrt3}{\frac{9+\sqrt3}{3}}\not=\frac{-29+9\sqrt3}{26}$

$tan\theta=\frac{-1+\sqrt3}{\frac{9+\sqrt3}{3}}=\frac{-3+3\sqrt3}{9 + \sqrt{3}}$

Nothing else jumped out at me, and unfortunately I do not have time to go over the whole thing more carefully. I hope that helped, though, and that someone will have a little more time than I do.

3. Originally Posted by Kevlar
Hi guys i've attempted this question 3 times now yet i keep getting the wrong answer so i turn to you my peers for help!.
Question:
Given that $cot(30°-\theta)=3$, find the value of $cot(\theta)$
My Workings Out:
Using $tan(30°-\theta)=\frac{tan(30°)-tan(\theta)}{1+tan(30°).tan(\theta)}$
$cot(30°-\theta)=\frac{1+tan(30°).tan(\theta)}{tan(30°)-tan(\theta)}$
$3=\frac{1+tan(30°).tan(\theta)}{tan(30°)-tan(\theta)}$
now i did tan30 which gives $\frac{\sqrt3}{3}$ i get
$3=\frac{1+\frac{\sqrt3}{3}tan(\theta)}{\frac{\sqrt 3}{3}-tan(\theta)}$
$3(\frac{\sqrt3}{3}-tan(\theta))=1+\frac{\sqrt3}{3}tan(\theta)$
${\color{red}\sqrt3-3tan(\theta)=1+\frac{\sqrt3}{3}tan\theta}$
re-arrange gives me
$-1+\sqrt3=\frac{9+\sqrt3}{3}tan(\theta)$
$tan\theta=\frac{-1+\sqrt3}{\frac{9+\sqrt3}{3}}=\frac{-29+9\sqrt3}{26}$
so
$cot\theta=\frac{26}{-26+9\sqrt3}$
$\frac{1+3\sqrt3}{3-\sqrt3}=\frac{6+5\sqrt3}{3}$
Hi Kevlar,

I'm going to pick it up at the red line above and finish a different way.

${\color{red}\sqrt3-3\tan \theta=1+\frac{\sqrt3}{3}\tan\theta}$

Multiply by 3

$3\sqrt{3}-3=\sqrt{3}\tan \theta+9 \tan \theta$

$3(\sqrt{3}-1)=\tan \theta(\sqrt{3}+9)$

$\tan \theta=\frac{3(\sqrt{3}-1)}{\sqrt{3}+9}$

$\cot \theta=\frac{\sqrt{3}+9}{3(\sqrt{3}-1)}\cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{3+10\sqrt{3}+9 }{3(3-1)}=\frac{12+10\sqrt{3}}{6}=\frac{6+5\sqrt{3}}{3}$