Need trigonometry help, simple problem?

Sin and cosine are periodic functions whose values vary between -1 and 1. Due to the pythogoreas theorm Sin**2X + Cos**2X = 1. So when SinX =0 CosX = 1.

10pi/3 = 3pi + pi/3. So this lies in the 3rd quadrant. In the 3rd quadrant the value of Sin is positive.

Sin10pi/3 = sinpi/3 = sqrt(3)/2. Cos10Pi/3 = -cospi/3 = -1/2.

Tan10pi/3 = Sin10pi/3/Cos10pi/3 = - sqrt(3).

pi is 180 degrees. So 10pi/3 is 600 degrees and then you subtract 360: 240.

The reference angle for 240 is 60 degrees.

The sin for 60 degrees is: (sqroot of 3)/2

The cos of 60 degrees is: 1/2

The tan for 60 degrees( sin over cosine): sqroot of 3

These shoulld be memorized by the way.

10/3 is a little over 3, so this is more than 2pi. Subtract 2pi because multiples of 2pi don't change anything.

10pi/3 - 2pi = 10pi/3 - 6pi/3 = 4pi/3

Now this is a little over pi, so we can write it as pi + pi/3. Now use various theorems about adding pi.

sin(pi + x) = -sin(x)

cos(pi + x) = -cos(x)

So in summary sin(10pi/3) = sin(4pi/3) = -sin(pi/3)

cos(10pi/3) = cos(4pi/3) = -cos(pi/3)

tan(10pi/3) = sin(10pi/3)/cos(10pi/3)

Convert given into numerical value:

(10 times 3.14)/3

314/3 = 104.7

Then use trig value tables to find equivalent angle.

http://www.sosmath.com/tables/trigtable/trigtable....

10π/3 is coterminal with 4π/3 (240°)

sin(4π/3) = –√(3)/2

cos(4π/3) = –1/2

tan(4π/3) = √(3)