Evaluating Trigonmetrix Functions

eamoss

Anyone what to show me how to do this with out a calculator?


For anyone who has the Thomas' Calculus book its page 56 question 6.

Fremen

If you're using Thomas' calculus, I guess you're at university level.
Remember from the leaving cert how to work out what sin and cos of 45, 30 and 60 are?

Remember how to convert degrees into into radians?

If you can work that out, you're halfway there.

Now find where each of the angles you're given are on the unit circle: the cosine is the first co-ordinate and the sine is the second co-ordinate.
You can remember it using the mnemonic (Christian name, Surname)->(Cos, Sin)
Remember, for negative angles you move clockwise from the x-axis.

If you need more hints, post back here.
This looks suspiciously like homework, so I'm a bit reluctant to do it for you.

eamoss

Fremen wrote: »

If you're using Thomas' calculus, I guess you're at university level.

If you need more hints, post back here.
This looks suspiciously like homework, so I'm a bit reluctant to do it for you.

You would be right on both, but the homework due date was Friday, 12 October 2007, 04:00 PM. Got 100% in it but that was because I had the solutions after that week they stopped setting questions from the book.

Can you do one step by step, I have notes on it but I dont understand it.

Fremen

ugh, this is going to be a pain in the arse without mathematical fonts.
Right, I'll try to explain the -PI/6 column.

Remember that PI radians = 180 degrees
then PI/6 radians = 30 degrees.

How do you work out what sin and cos of 30 are?
Draw an equilateral triangle with side length 2. The angles of this triangle are 60 degrees each.
Bisect one of the angles.
You have a right-angled triangle angles 30,60 and 90, and side-lengths 1, 2 and (by pythagoras' theorem) 1/(root 3).

By looking at this, you can find sin and cos of PI/6 and PI/3 radians.
See here:
http://www.themathpage.com/aTrig/30-60-90-triangle.htm

Now, if you want to find sin and cos of negative numbers, or numbers greater than PI/2 (90 degrees), you need to look at the unit circle. If you start at X=1, Y=0, and move anti-clockwise around the circle so that you make an angle THETA with the x-axis, then the cos of THETA is the x-coordinate and the sin is the y-coordinate of your point. This is the definition of sin and cos.

Hopefully, this will make it clear. And no, I can't use MS-paint very well.
http://img185.imageshack.us/img185/2167/triguj3.png

Now, it's just a matter of finding where your angles are on the unit circle, and reading off the coordinates to find Cos and Sin.
In the case of -pi/6, we know it's a 30-degree turn clockwise from the x-axis. We can see that the x-coordinate is the same as a 30-degree turn anticlockwise from the x-axis, i.e. cos(-pi/6) = cos(pi/6)

The Y coordinate is the same SIZE as it would be if we made a 30-degree turn anticlockwise, but it's the opposite SIGN, i.e. sin(-pi/6) = -sin(pi/6)

From the work earlier with the 30,60,90 triangle, you can see that cos(pi/6) = 1/2 and sin(pi/6) = 1/(root 3)

Then sin(-pi//6) = -1/(root 3) and cos(-pi/6) = cos(pi/6) = 1/2

The rest follow easily, for example cot(-pi/6) = cos(-pi/6) / sin(-pi//6) by definition, so its value is (1/2) / -(1/(root 3)) = -(root 3)/2

eamoss

Wow!

Thanks very much!!!

Fremen

No problem. I came back to london a week early for my college course by accident, so now I have f**k-all to do apart from messing around on the internet and drinking coffee

aequinoctium

there is a nice trick for when dealing with the sine, cosine (and thus all the others in your list) for the angles 0, 30, 45, 90

sine:
write out the numbers 0 to 4 (one below each of the angles).
divide each number by 4 and get the square root of the result...

0 1 2 3 4
0 1/4 1/2 3/4 1
0 1/2 1/[(2)^1/2] [(3)^1/2] / 2 1

which are the values of sin0 sin30 sin45 and sin90

cosine:
do the same but write the numbers from 4 to 0.

mp3guy

Doing a bit of study for MT101S I see

Fremen

Whoops, it would appear I f***ed up in my post:

side length of the 30-60-90 triangle is 1,2,(root 3), not 1/(root 3)

Also, sin(pi/6) is 1/2 which makes sin(-pi/6) -1/2
cos(pi/6) is (root3)/2 which makes cos(-pi/6) (root 3)/2

The method I outlined was sound but the numbers were wrong. Mostly because I'm an idiot.