Can somebody please help me explain this pre-calculus problem?

Sine and cosine are 2 trigonometric periodic functions, identical except that they are phase shifted. You can see that here: http://en.wikipedia.org/wiki/File:Sine_cosine_plot...

This phase shift can be written: sin(x+pi/2)=cos(x).

Proof:

We know that sin(a+b)=sin(a)cos(b)+sin(b)cos(a), so sin(x+pi/2)=sin(x)cos(pi/2)+sin(pi/2)cos(x); cos(pi/2)=0, sin(pi/2)=1 => sin(x+pi/2) = cos(x).

Also cos(x-pi/2) = sin(x)

Proof:

cos(a-b)=cos(a)cos(b)+sin(a)sin(b); cos(pi/2)=0, sin(pi/2)=1 => cos(x-pi/2)=sin(x)

Regarding calculus, well the derivative of the sine function is the cosine function and the derivative of the cosine function is -sine.

(sin(x))' = cos(x)

(cos(x))' = -sin(x)

The sine and cosine graphs both have maximum and minimum amplitudes at 1 and -1. The cosine graphs is an exact translation of the sine graph shited along 90degrees (pi/2; pi is 180degrees). That is at 0 degrees sin=0, cosin=1.

Sine and cosine are cofunctions of each other. Hence that, when A and B are complementary angles (that means when the two angles A and B add up to 90degrees, so A could be 10, B could be 80 and so they would be complementary) sinA = cosB.

Unfortunately for all of us, there is no fast & simple explanation. The basics are something like this:

A typical wave can be found in Alternating Current (AC) in your house.

Actually, Use the link below, it explains it far more logically than I can.

Look along the bottom of the cosine graph to get info on pi / 2

Can somebody please explain how to solve this problem: 61° 12' - 33° 37'. ---------------------------- Since, as you say, 1 degree = 60', you must do the equivalent of an improper fraction and rewrite the problem as: 60 deg 72' - 33 deg 37'. 60-33 = 27 deg and 72-37 = 35 '. put it all together for 27 deg 35'