What is the Simple Explanation for Radians and Trigonometric Functions?

[somebodyelse5]

Im currently in Calc II and am finding myself totally lost when it comes to solving things like sin(pi/3) and trig of that nature. I am very reliant on my calculator and am extremely fond of degrees, but I NEED to be able to find the definite integral of cos(x) from 0 to (pi/3) and problems that are similar. I was hoping that someone could help explain how this works. Its something I really should know how to do, but I don't know where else to go to learn.

Thanks in advance guys.

 

[Integral]

In general you will get more useful answers if you ask a specific question.

Can you narrow it down a bit? Just what is that you are having trouble with?

 

[somebodyelse5]

In general you will get more useful answers if you ask a specific question.

Can you narrow it down a bit? Just what is that you are having trouble with?

Sorry about that. Basically I have absolutely no idea how to evaluate an expression like sin(pi/3) without using my calculator. Would someone be able to explain to me how to do this?

I know that the answer is sqrt(3)/2 but I do not know how to do this without using my calculator.

 

[Bohrok]

You should memorize the values of sine, cosine, and tangent for common angles which you can see here
http://upload.wikimedia.org/wikipedia/commons/c/c9/Unit_circle_angles.svg

This should also be helpful for memorizing the values of sine and cosine in the first quadrant
http://en.wikipedia.org/wiki/Trig_i...monic_for_certain_values_of_sines_and_cosines

 

[somebodyelse5]

You should memorize the values of sine, cosine, and tangent for common angles which you can see here
http://upload.wikimedia.org/wikipedia/commons/c/c9/Unit_circle_angles.svg

This should also be helpful for memorizing the values of sine and cosine in the first quadrant
http://en.wikipedia.org/wiki/Trig_i...monic_for_certain_values_of_sines_and_cosines

Thanks! For the unit circle, the points (x,y)

does x correspond with cos and y correspond with sin? How does that part work?

 

[Bohrok]

Thanks! For the unit circle, the points (x,y)

does x correspond with cos and y correspond with sin? How does that part work?

Yes, that's right. Wikipedia can probably explain it better than I can for now (it's getting late here)
http://en.wikipedia.org/wiki/Trigonometric_functions

 

[Anonymous217]

Yes. Basically, it's SOH CAH TOA, in which case H, the hypotenuse, is 1 for unit circles. This makes sin(theta) dependent on the y value, cos(theta) dependent on the x value, and etc. I'm a bit curious why you've haven't learned this before calc.

 

[sponsoredwalk]

There's very little you need to do when thinking about these things luckily.

pi/3 is the same as 180°/3 = 60°.

I find it immensely easier to think in this way, so sin(60°)= (√3/2)

The reason I know this is from using SOH, CAH, TOA on the famous 30°, 60° & 90° triangles.
http://www.youtube.com/watch?v=Qwet4cIpnCM&feature=PlayList&p=26812DF9846578C3&playnext_from=PL

http://en.wikipedia.org/wiki/Special_right_triangles



\int_{0}^{ \frac{ \pi }{3}} cos(x)\,dx \ = \ sin( \frac{\pi}{3}) \ - \ sin(0) \ = \ \frac{ \sqrt{3} }{2}

Try doing it this way until you intuitively get that pi/3 = 60°, pi/4 = 45° etc... and don't be afraid to draw and redraw the 30°, 60°, 90° triangles in a margin so that you don't have to memorize everything, you can recall all of them with no hassle this way.

 

[HallsofIvy]

To find sin(30)= sin(\pi/6) or sin(60)= sin(\pi/3) and the cosines, think of an equilateral triangle. If you drop a perpendicular from one vertex to the opposite side, it bisects that opposite side. Now you have two right triangles with angles of 30 degrees= \pi/6 radians and 60 degrees= \pi/3 radians. If each side of the equilateral triangle, and the hypotenuse of each right triangle, was 1, the side opposite the 30 degree angle has length 1/2 and, by the Pythagorean theorem, the side opposite the 60 degree angle is \sqrt{3}/2. Now, you can calculate all trig functions from the definitions.

For 45 degrees= \pi/4, use an isosceles right triangle with legs of length 1. The hypotenuse has length \sqrt{2}.