# How do you find the exact values of cot, csc and sec for 45 degrees?

Jun 26, 2018

$\cot {45}^{\circ} = 1 \textcolor{w h i t e}{\frac{2}{2}} \csc {45}^{\circ} = \sqrt{2} \textcolor{w h i t e}{\frac{2}{2}} \sec {45}^{\circ} = \sqrt{2}$

#### Explanation:

Recall the definitions of the reciprocal trig identities:

$\cot x = \frac{1}{\textcolor{s t e e l b l u e}{\tan x}} \textcolor{w h i t e}{\frac{2}{2}} \csc x = \frac{1}{\textcolor{red}{\sin x}} \textcolor{w h i t e}{\frac{2}{2}} \sec x = \frac{1}{\textcolor{p u r p \le}{\cos x}}$

On the unit circle, the coordinates for ${45}^{\circ}$ are

$\left(\frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{2}\right)$

Where the $x$ coordinate is the $\cos$ value, and the $y$ coordinate is the $\sin$ value.

$\tan x$ is defined as $\sin \frac{x}{\cos} x$. From this information, we know

$\textcolor{s t e e l b l u e}{\tan x = 1} \textcolor{w h i t e}{\frac{2}{2}} \textcolor{red}{\sin x = \frac{\sqrt{2}}{2}} \textcolor{w h i t e}{\frac{2}{2}} \textcolor{p u r p \le}{\cos x = \frac{\sqrt{2}}{2}}$

We can plug these values into our expressions for the reciprocal identities to get

$\cot x = \frac{1}{\textcolor{s t e e l b l u e}{1}} = 1 \textcolor{w h i t e}{\frac{2}{2}} \textcolor{red}{\csc x = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}} \textcolor{w h i t e}{\frac{2}{2}} \textcolor{p u r p \le}{\sec x = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}}$

Remember for $\csc {45}^{\circ}$ and $\sec {45}^{\circ}$, we had to rationalize the denominator.

$\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \cot {45}^{\circ} = 1 \textcolor{w h i t e}{\frac{2}{2}} \csc {45}^{\circ} = \sqrt{2} \textcolor{w h i t e}{\frac{2}{2}} \sec {45}^{\circ} = \sqrt{2} \textcolor{w h i t e}{\frac{2}{2}} |}}$

Hope this helps!

Jun 27, 2018

$\cot {45}^{\circ} = \frac{1}{1}$
$\text{ } \csc {45}^{\circ} = \frac{\sqrt{2}}{1}$
$\text{ "" } \sec {45}^{\circ} = \frac{\sqrt{2}}{1}$

#### Explanation:

Draw a ${45}^{\circ}$ right triangle. A ${45}^{\circ}$ right triangle has 2 equal sides plus the hypotenuse, right? If you give the 2 equal sides a length of 1 unit, Pythagoras will tell you that the hypotenuse has a length of $\sqrt{2}$.

Label the lengths of the sides and the degrees of the angles. Keep this drawing handy.

You should have these memorized:
sine is opposite/hypotenuse

This question is asking for cot, csc and sec. Memorize these facts too:
cosecant is 1/sine
secant is 1/cosine
cotangent is 1/tangent

So thinking about the above things in bold,
$\cot {45}^{\circ} = \text{adjacent/opposite}$
$\csc {45}^{\circ} = \text{hypotenuse/opposite}$
$\sec {45}^{\circ} = \text{hypotenuse/adjacent}$

Now look at your drawing of the ${45}^{\circ}$ right triangle and plug in the lengths specified in the above fractions for each trig function.

I hope this helps,
Steve