please help, trig prob!?

Question

I am given sin 60 degree = (sq root 3) / 2, use the trig identities to find the exact values of

a) cos 30 degree

b) cos^2 60 degrees

c) sec (pi / 6)

d) csc (pi / 3)


please help and explain this to me, the book has a paragraph for an explain and does a crappy job at explaining it as well.

THANKS!

Answer 1

Question a)
cos 30° = √3 / 2

Question b)
cos² 60° = (1/2) x (1/2) = 1 / 4

Question c)
1 / cos (π / 6) = 1 / (√3 / 2) = 2 / √3 = (2√3 / 3 )

Question d)
1 / sin (π / 6) = 1 / (1 / 2) = 2

Answer 2

For the first and second you have to know about the cofunction properties. In that for any co-function (like sine and COsine or secant and COsecant) 90 degrees minus the angle is the angle for it's counter part:

sin 60 = cos 30 >>> sin (90 - 60) = cos 30

Therefore cos 30 would be root3 over 2. And for the second it would be 3/4.


For the third and fourth we need to know the definitions of sec and csc:

sec x = 1/(cos x)
csc x = 1/(sin x)

And how to convert from degrees to radians and back.

D>R >>> x * pi/180
R>D >>> x *180/pi

So:

sec (pi/6) = sec 30 = 1/(cos 30) = 1/(root3/2) = 2/root3 = (2root3)/3

csc (pi/3) = csc 60 = 1/(sin60) = 1/(root3/2) = 2/root3 = (2root3)/3

Answer 3

http://www.clarku.edu/~djoyce/trig/cosines.html

Answer 4

use a ti-89 duh