trig help please

[kannlowery]

Hi,
I'm taking a trig class at Purdue...I have a homework problem that says:
2sin60deg - sec45deg tan60deg=?
The answer in the back of the book says : sq root3 - sq root6 (sorry I don't have some symbols on my computer). How did they get this?

 

[tkhunny]

Hi,
I'm taking a trig class at Purdue...I have a homework problem that says:
2sin60deg - sec45deg tan60deg=?
The answer in the back of the book says : sq root3 - sq root6 (sorry I don't have some symbols on my computer). How did they get this?

There are a few values you should memorize.

These are some of them:
sin(60°) = \(\displaystyle \frac{\sqrt{3}}{2}\)
cos(45°) = sin(45°) = \(\displaystyle \frac{\sqrt{2}}{2}\)
cos(60°) = 1/2

That should get you through this problem.

 

[Guest]

2sin60deg - sec45deg tan60deg=?

I assume at some stage you have drawn a 60,30,90 degree triangle with sides of 1,sqrt 3, and 2.
As well a 45,45,90 triangle with sides of 1,1, sqrt 2

then the conversions of sin 60 = (sqrt 3 )/ 2
sec 45 = (sqrt 2) / 1
tan 60 = (sqrt 3) / 1

place these into your above equation, simply and you are finished.