1. ## Quick sin,cos,tan question

Given tan(theta) = 8, find cos(theta) and sin(theta)

I have no idea how to do this lol, i forgot >_< and i want exact answers plz >_< so no decimals 0_o

2. think of the right triangle.
$\cos\theta=\frac{adjacent}{hypotenuse}$
$\sin\theta=\frac{opposite}{hypotenuse}$
$\tan\theta=\frac{opposite}{adjacent}$
Also Pythagorus' theorem: $a^2=b^2+c^2$
where a is the hypotenuse, and b and c are adjacent and opposite sides respectively.

3. Essentially the same: think of $tan(\theta)= 8$ as representing a right triangle with "near side" of length 8 and "opposite side" of length 1 so that $tan(\theta)$= "near side"/"opposite side"= 8. Use the Pythagorean theorem to find the hypotenuse and then use the formulas arze gives.

Given tan(theta) = 8, find cos(theta) and sin(theta)

I have no idea how to do this lol, i forgot >_< and i want exact answers plz >_< so no decimals 0_o
You haven't defined your domain, so you may also consider $Tan\theta$ to be the slope of a line through the origin, for $0\ \le\ \theta\ \le\ 2{\pi}$.

Hence, there are also negative solutions for $Sin\theta$ and $Cos\theta$.

You don't know the lengths of the sides of your right-angled triangle, but all you need is the ratios.

$Adjacent=x$

$Opposite=8x$

$Hypotenuse=\sqrt{(8x)^2+x^2}=x\sqrt{65}$

$\text{Given: }\:\tan\theta \,=\, 8$

$\text{Find: }\;\cos\theta\,\text{ and }\,\sin\theta$

I will assume that $\,\theta$ is an acute angle.

$\text{We have: }\;\tan\theta \:=\:\dfrac{8}{1} \:=\:\dfrac{opp}{adj}$

That is: . $\,\theta$ is in a right triangle with $opp = 8,\;adj = 1$
. .
(Can you picture that? If not, make a sketch.)

Pythagorus tell us that: . $hyp \:=\:\sqrt{8^2+1^2} \:=\:\sqrt{65}$

Therefore: . $\begin{Bmatrix}\cos\theta &=& \dfrac{adj}{hyp} &=& \dfrac{1}{\sqrt{65}} \\ \\[-3mm] \sin\theta &=& \dfrac{opp}{hyp} &=& \dfrac{8}{\sqrt{65}} \end{Bmatrix}$

Given tan(theta) = 8, find cos(theta) and sin(theta)

I have no idea how to do this lol, i forgot >_< and i want exact answers plz >_< so no decimals 0_o
Here is a different approach.
$
\tan\theta= \frac{\sin\theta}{\cos\theta} =8$

$\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=64$

$\sin^2\theta=64 \cos^2\theta$

$\sin^2\theta =64 (1-\sin^2\theta)$

$\sin^2\theta =64-64\sin^2\theta$

$65\sin^2\theta = 64$

$\sin^2\theta = \frac{64}{65}$

$\sin\theta = \frac{8}{\sqrt{65}}$

$
\tan\theta= \frac{\sin\theta}{\cos\theta} =8$

$\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=64$

$\frac{1-\cos^2\theta}{\cos^2\theta}=64$
$
\frac{1}{\cos^2\theta}-\frac{\cos^2\theta}{\cos^2\theta}=64$

$\frac{1}{\cos^2\theta}-1=64$

$\frac{1}{\cos^2\theta}=65$

$\cos^2\theta=\frac{1}{65}$

$\cos\theta=\frac{1}{\sqrt{65}}$