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Reference Angles

0
1999
2

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?

Apr 26, 2015

#1
+118627
+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

$$\\cos (\pi/4) = cos 45^0 = \frac{1}{\sqrt2}=\frac{\sqrt2}{2}=\frac{\frac{\sqrt2}{2}}{1}=\frac{adj}{hyp}\\\\ the hypotenuse is 1 because it is the radius of the unit circle\\ so the x value must be \frac{\sqrt2}{2} but it is negative, you can see that the x is negative there!\\ now \\ sec\left\frac{\pi}{4}\right=\frac{hyp}{adj}=\frac{1}{x\;co-ordinate}=\frac{1}{-\frac{\sqrt2}{2}}=\frac{-2}{\sqrt2}$$

Apr 26, 2015