a) sec(3pi/4) reference angle: pi/4 and Value is -2/sqroot2

but how?

b)cot(-2pi/3) reference angle: pi/3 and Value is 1/ sqroot 3

but how?

Guest Apr 26, 2015

#1**+10 **

okay part a

a) sec(3pi/4) reference angle: pi/4 and Value is -2/sqroot2

I have drawn a unit circle. It is an excellent idea for you to completely understand the unit circle and its significance. It makes much of trigonometry much easier. :)

A unit circle is a circle centre (0,0) and radius 1 unit.

I have drawn the 3pi/4 angle where it belongs. If you drop a perpendicular from the ray down onto the x axis you will form the reference angle at at the origin. So the reference angle is pi/4

NOW sec is 1/cos

$$\\cos (\pi/4) = cos 45^0 = \frac{1}{\sqrt2}=\frac{\sqrt2}{2}=\frac{\frac{\sqrt2}{2}}{1}=\frac{adj}{hyp}\\\\

$the hypotenuse is 1 because it is the radius of the unit circle$\\

$so the x value must be $\frac{\sqrt2}{2}$ but it is negative, you can see that the x is negative there!\\

now \\

sec\left\frac{\pi}{4}\right=\frac{hyp}{adj}=\frac{1}{x\;co-ordinate}=\frac{1}{-\frac{\sqrt2}{2}}=\frac{-2}{\sqrt2}$$

Melody Apr 26, 2015