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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
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 #1
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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

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