Tuesday, August 31, 2010

Charlie's Blog #2

A.) Cosine and sine have a relationship in which, using the Pythagorean Theorem, sinΘ + cosΘ = 1. In any angle using the theorem they will always equal 1. [http://www.math.dartmouth.edu/opencalc2/cole/lecture9.pdf]
Also, on the trigonometry chart, sine is the opposite order of cosine. All the numbers and factions are the same, just opposite.
*Sin 0 = 0 Cos 0 = 1
Sin pi/6 = ½ Cos pi/6 = square root of 3/2
Sin pi/4 = square root of 2/2 Cos pi/4 = square root of 2/2
Sin pi/3 = square root of 3/2 Cos pi/3 = 1/2
Sin pi/2 = 1 Cos pi/2 = 0

B.) Tangent and Cotangent are in relation with there values. Where tan = y/x and cot = x/y. So if tan equals 4/9 then cot equals 9/4.
There is also the relationship on the trigonometry chart. Like sine and cosine, these are also opposite of each other.
*Tan 0 = 0 Cot 0 = 0
Tan pi/6 = square root of 3/3 Cot pi/6 = square root of 3
Tan pi/4 = 1 Cot pi/4 = 1
Tan pi/3 = square root of 3 Cot pi/3 = square root of 3/3
Tan pi/2 = undefined Cot pi/2 = 0

C.) Sine and Cosecant relate in the way that sin = y/r and csc = r/y.
So on the ‘Trig Chart’, sine flipped equals cosecant.
*Csc 0 = undefined -instead of 0
Csc pi/6 = 2 -instead of ½
Csc pi/4 = square root of 2 -instead of square root of 2/2
Csc pi/3 = 2 square root of 3/2 -instead of square root of 3/2
Csc pi/2 = 1 -the same as sine

D.) Secant and Cosine are related in the way that cos = x/r while sec = r/x.
And on the trigonometry chart, like sine and cosecant, it’s cosine flipped.
*Sec 0 = 1 -the same as cosine
Sec pi/6 = 2 square root of 3/3 -instead of square root of 3/2
Sec pi/4 = square root of 2 -instead of square root of 2/2
Sec pi/3 = 2 -instead of ½
Sec pi/2 = undefined -instead of 0

D.) Sine, Cosine, and tangent are alike because tangent can be found if given sine and/or cosine. Tangent equals y/x, sine equals y/r, and cosine equals x/r. Therefore if sinΘ = 3/6 and cosΘ = 9/6, then tan = 3/9.

1 comment:

  1. Charlie,

    Great work on the blog but post it as a comment next time.

    ReplyDelete