1. ## Trigonometry Help [Q1]

Question:
Given $\displaystyle x = \arcsin \frac{1}{3}$ , find the exact value of:
a. $\displaystyle \cos x$
b. $\displaystyle \tan x$
c. $\displaystyle \cot x$
d. $\displaystyle \sec x$
e. $\displaystyle \csc x$

Attempt:

a. $\displaystyle \cos x$

$\displaystyle = \cos (x = \arcsin \frac{1}{3})$

$\displaystyle = 0.942809041$

$\displaystyle \color{red}{= 2 \frac{\sqrt{2}}{3}}$

How can I get the fraction answer?

2. Originally Posted by looi76
Question:
Given $\displaystyle x = \arcsin \frac{1}{3}$ , find the exact value of:
a. $\displaystyle \cos x$
b. $\displaystyle \tan x$
c. $\displaystyle \cot x$
d. $\displaystyle \sec x$
e. $\displaystyle \csc x$

Attempt:

a. $\displaystyle \cos x$

$\displaystyle = \cos (x = \arcsin \frac{1}{3})$

$\displaystyle = 0.942809041$

$\displaystyle \color{red}{= 2 \frac{\sqrt{2}}{3}}$

How can I get the fraction answer?
$\displaystyle x= \sin^{-1}\left(\frac{1}{3}\right)$

$\displaystyle \Rightarrow \sin x =\frac{1}{3}$

$\displaystyle \cos x = \sqrt{1-\sin^2 x}=\sqrt{1-\left(\frac{1}{3}\right)^2 }=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$

$\displaystyle \tan x = \frac{\sin x}{\cos x}=\frac{.......}{.......}$ finish it.

$\displaystyle \cot x = \frac{1}{\tan x}=\frac{1}{.......}$ finish it.

$\displaystyle \sec x = \frac{1}{\cos x}=\frac{1}{.......}$ finish it.

$\displaystyle \csc x = \frac{1}{\sin x}=\frac{1}{.......}$ finish it.

got it ???

3. Originally Posted by Shyam
$\displaystyle x= \sin^{-1}\left(\frac{1}{3}\right)$

$\displaystyle \Rightarrow \sin x =\frac{1}{3}$

$\displaystyle \cos x = \sqrt{1-\sin^2 x}=\sqrt{1-\left(\frac{1}{3}\right)^2 }=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$

$\displaystyle \tan x = \frac{\sin x}{\cos x}=\frac{.......}{.......}$ finish it.

$\displaystyle \cot x = \frac{1}{\tan x}=\frac{1}{.......}$ finish it.

$\displaystyle \sec x = \frac{1}{\cos x}=\frac{1}{.......}$ finish it.

$\displaystyle \csc x = \frac{1}{\sin x}=\frac{1}{.......}$ finish it.

got it ???
$\displaystyle \cos x = \pm \sqrt{1-\sin^2 x}=\pm \sqrt{1-\left(\frac{1}{3}\right)^2 }=\pm \sqrt{\frac{8}{9}}=\pm \frac{2\sqrt{2}}{3}$

CB