Trig. Maths Q.

[danza]

I can't work out what I'm meant to do here. I've forgotten loads of the maths I did last year at college, and from what I remember, we missed this anyway.

Here's one of the questions:

Write expressions for cos(30°), sin(30°). Convert them to numerical values.

I know this is simple, but I really can't remember what to do.

AFAIK, the expression for cos(30°) is sqrt(3)/2, but I can't see how/why? Can someone help me get my mind back in gear?

 

[dirtychinchilla]

well the reason why it's expressed as sqrt3/2 is because cos 30 isn't a integer, ie is not a whole number. i don't actually know how to convert it correctly though...it's a value that you just know. and sin30 is 1/2

 

[RomanNose]

I think you can deduce using pythagoras triangle. Just figuring it out as I type.

Ok take an equilateral triangle with sides of equal length of two, with lengths BC CA and BA
Now bisect BC into two, at point D halfway between BC

Now you can then use pythogoras as AB^2= BD^2 + DA^2
So 4=1+Ab^2
So AB=Sqrt 3
Then just use cos 30 = a/h
so cos 30 = Sqrt 3/2.

 

[danza]

Cheers for that, I understand what you've done there no problem. But I also have to write expressions for cos(45) (which I know to be 1/sqrt(2)), sin (45), and another question for cos(60), sin(60).

I really can't think what I'm meant to do (maybe I'm missing something blatently obvious). It's not 'homework', just an exercise I need to do. Well, more need to understand than anything really.

 

[mufc802]

You can also draw a right angled triangle to find cos sin tan45. Make both sides 1 and the longest side is sqrt2. sin45 and cos45 are both 1/sqrt2 which is the same as sqrt2/2, tan45 is just 1.

 

[danza]

Yep, it clicked whilst I was eating my bolognese.

Cheers guys.