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MATH 1113 PreCalculus

Day #27, Trig Identities II

Circle Identities:

The first thing you should check is that the right triangle identities are preserved when we extend our definitions to arbitrary angles (using the circle definitions for the trig functions.)  (They are....)

But, in addition, once we've extended our definitions, we have several more fundamental relationships that the trig functions have, given the nature of the unit circle and our definitions.

The first and most obvious are the "periodicity identities," that come from the fact that every time you go around the circle (2*pi) you're back where you started from:

Periodicity Identities: for "k" any integer,

 

sin(t +/- 2*k*pi) = sin(t)  

csc(t +/- 2*k*pi) = csc(t)

cos(t +/- 2*k*pi) = cos(t)  

sec(t +/- 2*k*pi) = sec(t)

tan(t +/- k*pi) = tan(t)*  

cot(t +/- k*pi) = cot(t)*

*notice that tangent and cotangent have period pi, due to the first "shift" identity, below.

The "shift" identities: Notice that if you go 180 degrees (pi radians) from any point (x,y) on the unit circle, you end up at (-x, -y).  Since sin(t) = y and cos(t) = x, we get, automatically:

Shift identities (I):

sin(t +/- pi) = -sin(t)  

csc(t+/- pi) = -csc(t)

cos(t +/- pi) = -cos(t)  

sec(t +/- pi) = -sec(t)

tan(t +/- pi) = tan(t)**  

cot(t +/- pi) = cot(t)**

**tan(t +/- pi) = sin(t +/- pi)/cos(t +/- pi) = -sin(t)/(-cos(t)) = sin(t)/cos(t) = tan(t) -- the two negatives cancel out.  The same holds for the cotangent, thus proving the * below the first table.

A more complicated (but important) shift relates the sine and cosine functions.  If you subtract pi/2 -- 90o from any angle, you'll see that the x-coordinate of the new angle is the same as the y-coordinate of the original angle -- see the Geogebra applet below:

Applet

This gives further identities relating sine and cosine:

Shift identities II:

sin(t - pi/2) = cos(t)  

cos(t + pi/2) = sin(t)

csc(t - pi/2) = sec(t)  

sec(t + pi/2) = csc(t)

tan(t +/- pi/2) = -cot(t) ***  

cot(t +/- pi/2) = -tan(t)***

***tan(t - pi/2) = sin(t - pi/2)/cos(t - pi/2)
                       = sin(t - pi/2)/(-cos(t + pi/2)) (from the first shift identity)
                       = cos(t)/(-sin(t)) = -cot(t)

As always, the tangent and cotangent functions have slightly different identities, because of the interplay of sines and cosines.

And finally, we get the symmetry identities, based on our way of defining the angle as working off of the positive x-axis, namely that:

sin(-t) = -sin(t)  

csc(-t) = -csc(t)

cos(-t) = cos(t)  

sec(-t) = sec(t)

tan(-t) = -tan(t)  

cot(-t) = -cot(t)

(Notice that here the cosine -- and secant -- are the different ones.)

Said another way: cosine and secant are EVEN functions, while the others are ODD functions.  It is also said that "Sines pass minus signs, while cosines absorb minus signs."

Here's a Geogebra applet demonstrating this symmetry:

Applet

Next, we'll look at how to prove some identities, concentrating on the right triangle ones.

 


For Practice, try p. 591 #7 - 13 odd, #25 - 31 odd, 59 - 69 odd

 

 


Assignment: Do the CourseCompass problems from section 7.1
 

 

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Page Last Modified: 22 March, 2007

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