1. ## arccos+arccot

Calculate . The answer may not contain cyklometriska functions.

got no ide to do this

2. ## Re: arccos+arccot

Hello, Petrus!

$\text{Calculate: }\:\theta \;=\;\underbrace{\arccos\left(\tfrac{3}{5}\right)} _{\alpha} + \underbrace{\text{arccot}\left(\tfrac{1}{7}\right) }_{\beta}$

Let $\alpha \,=\,\arccos\left(\tfrac{3}{5}\right) \quad\Rightarrow\quad \cos\alpha \,=\,\tfrac{3}{5} \,=\,\tfrac{adj}{hyp}$
. . Then: . $opp \,=\,4\quad\Rightarrow\quad \tan\alpha \,=\,\tfrac{4}{3}$

Let $\beta \,=\,\text{arccot}\left(\tfrac{1}{7}\right) \quad\Rightarrow\quad \cot\beta \,=\,\tfrac{1}{7}$
. . Then: . $\tan\beta \,=\,\tfrac{7}{1} \,=\,7$

We have: . $\theta \;=\;\alpha + \beta$

. . $\tan\theta \;=\;\frac{\tan\alpha + \tan\beta}{1 - (\tan\alpha)(\tan\beta)} \;=\;\frac{\frac{4}{3}+7}{1-\frac{4}{3}(7)} \;=\;\frac{\frac{25}{3}}{\text{-}\frac{25}{3}} \;=\;-1$

$\text{Therefore: }\:\theta \;=\;\frac{3\pi}{4} + \pi n\;\text{ for any integer }n.$

3. ## Re: arccos+arccot

Let $\mbox{arc}\cos\left(\frac{3}{5}\right)+\mbox{arc} \cot\left(\frac{1}{7}\right)=x$
Now, let's take the $\cos$ of both sides. Thus
$\cos\left[\mbox{arc}\cos\left(\frac{3}{5}\right)+\mbox{arc} \cot\left(\frac{1}{7}\right)\right]=\cos(x)$
$\Leftrightarrow \cos\left[\mbox{arc}\cos\left(\frac{3}{5}\right)\right]\cos\left[\mbox{arc} \cot\left(\frac{1}{7}\right)\right]-\sin\left[\mbox{arc}\cos\left(\frac{3}{5}\right)\right]\sin\left[\mbox{arc} \cot\left(\frac{1}{7}\right)\right]=\cos(x)$

If we use the following identities:
$\cos\left[\mbox{arc}\cos(x)\right]=x$
$\cos\left[\mbox{arc}\cot(x)\right]=\frac{1}{\sqrt{\left(\frac{1}{x}\right)^2+1}}$
$\sin\left[\mbox{arc}\cos(x)\right]=\sqrt{1-x^2}$
$\sin\left[\mbox{arc}\cot(x)\right] =\frac{1}{\sqrt{1+x^2}}$

then we can write the equation as
$\left(\frac{3}{5}\right)\frac{1}{\sqrt{50}}-\sqrt{1-\left(\frac{3}{5}\right)^2}\frac{1}{\sqrt{1+\left( \frac{1}{7}\right)^2}}=\cos(x)$
$\Leftrightarrow \frac{3}{5\sqrt{50}} - \frac{4}{5}\left(\frac{7}{\sqrt{50}}\right)=\cos(x )$
$\Leftrightarrow \frac{3}{25\sqrt{2}} - \frac{28}{25\sqrt{2}}=\cos(x)$
$\Leftrightarrow \cos(x) = \frac{-\sqrt{2}}{2}$
$\Leftrightarrow x = \frac{3\pi}{4}+2k\pi$

4. ## Re: arccos+arccot

Originally Posted by Soroban
Hello, Petrus!

Let $\alpha \,=\,\arccos\left(\tfrac{3}{5}\right) \quad\Rightarrow\quad \cos\alpha \,=\,\tfrac{3}{5} \,=\,\tfrac{adj}{hyp}$
. . Then: . $opp \,=\,4\quad\Rightarrow\quad \tan\alpha \,=\,\tfrac{4}{3}$

Let $\beta \,=\,\text{arccot}\left(\tfrac{1}{7}\right) \quad\Rightarrow\quad \cot\beta \,=\,\tfrac{1}{7}$
. . Then: . $\tan\beta \,=\,\tfrac{7}{1} \,=\,7$

We have: . $\theta \;=\;\alpha + \beta$

. . $\tan\theta \;=\;\frac{\tan\alpha + \tan\beta}{1 - (\tan\alpha)(\tan\beta)} \;=\;\frac{\frac{4}{3}+7}{1-\frac{4}{3}(7)} \;=\;\frac{\frac{25}{3}}{\text{-}\frac{25}{3}} \;=\;-1$

$\text{Therefore: }\:\theta \;=\;\frac{3\pi}{4} + \pi n\;\text{ for any integer }n.$

Was it necessary to use $\tan\theta$? I tried using $\sin\theta$, but I ended up getting $\sin\frac{5}{\sqrt(50)}$

5. ## Re: arccos+arccot

Hello soroban!
Idk where u got that formel tan o=tan a+tan b/... Etc
Then i wanna ask how do i see what tan o=-1 ik its writen on My book but on test i cant use My book so plz can some1 tell me? Ik for 30degree,60 and 45 i can make a triangel but how i do with rest!!!