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Notes on the Riding on the Where’s the Identity Learning Task
This task wraps up the unit by introducing common techniques used in establishing more
complex identities. Students often confuse establishing identities with solving equations, because
students are so used to solving an equation when given one. It is important to make sure
students understand that when asked to establish an identity, they are not solving an equation,
and that with an identity, any value of a variable for which the expressions in the identity are
defined will be a solution.
The first two problems address the common mistakes that can occur if students treat
establishing identities as problems similar to solving equations. Students then are given three
sets of identities to establish, with each set requiring the application of a specified group of the
basic identities listed in standard MM4A5. This task is a culmination of previous units, as it
requires knowledge of all identities learned in previous units.
This task addresses standard MM4A5.
ESTABLISHING IDENTITIES
As you saw in Where’s the Identity?, trigonometric identities can be difficult to recognize,
but by thinking graphically, numerically, and algebraically, you can gain valuable insight as to
whether equations are identities or not. Numerical and graphical information is not enough to
verify an identity, however.
Identities can be established algebraically by rewriting one side of the equation until it
matches the expression on the other side of the equation. Rewriting is often done by applying a
basic trigonometric identity, so the identities you have already established are listed here for
reference.
Quotient Identities
Reciprocal Identities
tan q =
sin q
cos q
cos q =
1
sec q
secq =
1
cos q
cot q =
cos q
sin q
sin q =
1
csc q
csc q =
1
sin q
tan q =
1
cot q
cot q =
1
tan q
Pythagorean Identities
sin 2 q + cos2 q = 1
tan 2 q + 1 = sec2 q
1 + cot 2 q = csc2 q
sin 2 q = 1 - cos2 q
tan 2 q = sec2 q - 1
1 = csc2 q - cot 2 q
cos2 q = 1 - sin 2 q
1 = sec2 q - tan 2 q
cot 2 q = csc2 q - 1
Sum & Difference Identities
Double Angle Identities
sin (a ± b ) = sin a cos b ± cos a sin b
sin (2q ) = 2sin q cos q
cos (2q ) = cos2 q - sin 2 q
When establishing identities, it is important that each equation that you write is logically
equivalent to the equation that precedes it. One way to ensure that all of your equations are
equivalent is to work with each side of the equation independently. The following problem
provides an example of how failing to work with each side of an equation independently can
produce what appears to be a proof of a statement that isn’t true. This example should serve as a
reminder as to why you should work with each side of an equation independently when
establishing identities.
1. As explained above, a helpful guideline when establishing identities is to change each side
of the equation independently. Circle the two lines that were produced by failing to abide
by this guideline, in the “faulty proof” below.
sin q = - 1 - cos2 q
sin 2 q = 1 - cos2 q
sin 2 q + cos2 q = 1
1=1
2. Explain why sin q = - 1 - cos 2 q is not an identity, using either graphical or numerical
reasoning.
Solution: Making a table of values that shows numeric differences between the expressions
of each side of the equation provides numerical reasoning as to why the equation is not an
identity. Although some values of  may produce equal values, for the statement to be an
identity, each side of the equation must be equal for all values of  , which is not the case in
this example:
sin 

 1  cos2 
 4
0
 22
0

4
2
2
 22
0
 22
Graphing each side of the equation as a function provides graphical reasoning as to why the
equation is not an identity:
f    sin 
f     1  cos2 
When establishing the following identities, keep the following two general rules of thumb in
mind. They will not always lead to the most efficient solution, but they are usually beneficial when
help is needed.


Begin working on the most complex side, because it is usually easier to simplify an
expression rather than make it more complex.
When no other solution presents itself, rewrite both sides in terms of sines and cosines.
Establish the following identities by rewriting the left, right, or both sides of the equation
independently, until both sides are identical.
For problems 3-5, apply either the quotient or reciprocal identities.
3. sin x cot x  cos x
Solution:
cos x
sin x
 cos x
sin x
cos x  cos x
4. csc tan  cos  1
Solution:
1 sin 

 cos   1
sin  cos 
11
sec 
 tan 
csc 
Solution:
5.
1
cos 
1
sin 
 tan 
 sin1  tan 
tan   tan 
1
cos
For problems 6-10, apply the Pythagorean Identities.
6. tan 2 w  8  sec2 w  7
Solution:
tan 2 w  8  1  tan 2 w  7
tan 2 w  8  tan 2 w  8
7. 4cos2 q + 3sin 2 q = 3 + cos2 q
Solution:
4cos 2   3 1  cos 2    3  cos 2 
4cos 2   3  3cos 2   3  cos 2 
3  1cos 2   3  cos 2 
tan 2 q - sin 2 q = tan 2 q sin 2 q
Solution:
tan 2   sin 2    sec 2   1 sin 2 
tan 2   sin 2   sec2  sin 2   sin 2 
sin 2 
tan 2   sin 2  
 sin 2 
2
cos 
2
2
tan   sin   tan 2   sin 2 
8.
sin  tan 
 sec   1
1  cos 
Solution:
sin  sin 

 sec  1
1  cos  cos 
sin 2 
 sec  1
cos 1  cos  
1  cos 2 
 sec  1
cos 1  cos  
1  cos 1  cos   sec  1
cos 1  cos 
1  cos
 sec  1
cos
1
cos 

 sec  1
cos cos
sec 1  sec 1
sec2   1
9.
 sin 2 
2
sec 
Solution:
sec2 
1

 sin 2 
2
2
sec  sec 
1  cos 2   sin 2 
sin 2   sin 2 
For problems 10-12, apply the sum, difference, or double angle identities.
cos    
10.
 cot   tan 
cos  sin 
Solution:
cos  cos   sin  sin 
 cot   tan 
cos  sin 
cos  cos  sin  sin 

 cot   tan 
cos  sin  cos  sin 
cos  sin 

 cot   tan 
sin  cos 
cot   tan   cot   tan 
11.
sin (a - b )
sin a cos b
Solution:
= 1 - cot a tan b
sin  cos   cos  sin 
 1  cot  tan 
sin  cos 
sin  cos  cos  sin 

 1  cot  tan 
sin  cos  sin  cos 
1  cot  tan   1  cot  tan 
12. 2cot x cot (2 x ) = cot 2 x - 1
Solution:
2cos x cos  2 x 

 cot 2 x  1
sin x sin  2 x 
2cos x cos 2 x  sin 2 x

 cot 2 x  1
sin x
2cos x sin x
2
cos x  sin 2 x
 cot 2 x  1
sin 2 x
cos 2 x sin 2 x
 2  cot 2 x  1
2
sin x sin x
cot 2 x  1  cot 2 x  1
13. cos4 q - sin 4 q = cos (2q )
Solution:
 cos   sin   cos   sin    cos  2 
2
2
2
1 cos  2   cos  2 
cos  2   cos  2 
2