Download Fundamental Trigonometric Identities Reciprocal Identities sin A = 1

Fundamental Trigonometric Identities
Reciprocal Identities
sin A =
cos A =
1
cos ecA
cosec A =
sec A =
1
s in A
tan A =
1
sec A
cot A =
1
cos A
1
cot A
1
ta n A
Quotient Identities
tan A =
cot A =
s in A
cos A
Pythagorean Identities
sin2 A + cos2 A = 1
Co-function Identities
 A
2
1 + cot2 A = cosec2 A
cos(   A ) = sin A
2
tan(
s in A
1 + tan2 A = sec2 A
sin(   A ) = cos A

cos A
2
) = cot A
cot(
sec(   A ) = cosec A

 A
2
) =
tan A
cosec(   A ) = sec A
2
2
Even/Odd Identities
sin(– A) = – sin A
cos(– A) = cos A
cosec(– A) = – cosec A sec(– A) = sec A
Example 1 (Using Identities to Evaluate a Function)
Use the values sec A =

3
2
tan(– A) = – tan A
cot(– A) = – cot A
and tan A > 0 to find the values of all six trigonometric
functions.
Solution
Using a reciprocal identity,
cos A =
1
sec A
=
1
3 / 2
= 2
3
Using a Pythagorean Identity
2
sin2A = 1 – cos2A = 1 –   2  = 1 –

3 
4
9
=
5
9
.
Because sec A < 0 and tan A > 0, it fallows that A lies in Quadrant III, then sin A is negative and
obtain sin A =

5
3
. Now, knowing the values of the sine and cosine, you can find the values of
all six trigonometric functions.
sin A =

5
3
cosec A =
1
s in A
=
3
5
cos A =
tan A =

sec A =
2
3
s in A
cos A
=

5 /3
2 / 3
=
cot A =
5
2
1
co s A
1
ta n A
=
=

3
2
2
5
Ans.
Example 2 (Simplifying a Trigonometric Expression)
Simplify sin x cos2 x – sin x
Solution
Factor the expression and then use a fundamental identity.
sin x cos2 x – sin x = sin x(cos2 x – 1)
= – sin x(1 – cos2 x)
= – sin x(sin2 x)
= – sin3 x
Ans.
Example 3 (Factoring Trigonometric Expressions)
Factor each expression.
a. s e c   1
b. 4 ta n   ta n   3
Solution
a. Here you have the difference of two squares, which factors as
sec   1 =  s e c   1  s e c   1
b. This expression has the polynomial form ax2 + bx + c, and it factors as
4 ta n   ta n   3 =  4 t a n   3   t a n   1 
Ans.
2
2
2
2
Example 4 (Factoring a Trigonometric Expression)
Factor cosec2 x – cot x – 3.
Solution
You can use the identity cosec2 x = 1 + cot2 x to rewrite the expression in term of the cotangent.
cosec2 x – cot x – 3 = (1 + cot2 x ) – cot x – 3
= cot2 x – cot x – 2
= (cot x – 2)( cot x + 1)
Ans.
Example 5 (Simplifying a Trigonometric Expression)
Simplify sin t + cot t cos t
Solution
Begin by rewriting cot t in terms of sine and cosine.
sin t + cot t cos t = sin t +  c o s t  cos t
 s in t 
=
=
s in t  c o s t
2
2
s in t
1
s in t
= cosec t
Ans.
Example 6 (Adding Trigonometric Expressions)
Perform the addition and simplify.
s in 
1  cos 
+
cos 
+
cos 
s in 
Solution
s in 
1  cos 
s in 
=
 s in    s in     c o s   1 
=
s in   c o s   c o s 
=
=
1 
2
cos  
c o s    s in  
2
1 
c o s    s in  
1  cos 
1 
c o s    s in  
1
s in 
= c o s e c
Ans.
Example 7 (Rewriting a Trigonometric Expression)
Rewrite
1
1  s in x
so that it is not in fraction form.
Solution
From the Pythagorean Identity cos2 x = 1 – sin2 x = (1 – sin x)(1+ sin x), you can see that by
multiplying both the numerator and the denominator by (1 – sin x) you produce a monomial
denominator.
1
1  s in x
=
=
1
1  s in x
1  s in x
1  s in x
2
= 1  s in x
2
cos x
. 1  s in x
1  s in x
=
=
1
2
cos x
–
1
2
cos x
s in x
2
cos x
–
s in x
1
.
cos x cos x
= sec2 x – tan x sec x
Ans.
Example 8 (Trigonometric Substitution)
Use the substitution x = 2 ta n  , 0 <  <  , to express
2
4  x
2
as a trigonometric
function of 
Solution
Begin by letting x = 2 ta n  . Then, you can obtain
4  x
= 4   2 ta n  
= 4  4 ta n 
= 4  1  ta n  
2
2
2
2
=
=
4 sec 
2
Ans.
2 sec 
Checking
4  x
2
x

2
FIGURE 5.1
Figure 5.1 shows the right triangle illustration of the trigonometric substitution in example
8. You can use this triangle to check the solution of example 8. For 0 <  <  , you have
2
tan x =
x
2
opp = x, adj = 2, and hyp = 4  x
With these expressions, you can write the following.
2
sec  =
h yp
adj
sec  =
4  x
2
2
2sec  = 4  x
So, the solution checks.
2
Guidelines for Verifying Trigonometric Identities
2.1. Work with one side of the equation at a time. It is often better to work with
the more complicated side first.
2.2. Look for opportunities to factor an expression, add fractions, square a
binomial, or create a monomial denominator.
2.3. Look for opportunities to use the fundamental identities. Note which
functions are in the final expression you want. Sines and cosines pair up well, as do secants
and tangents, and cosecants and cotangents.
2.4. If the preceding guidelines do not help, try converting all terms to sines and
cosines.
2.5. Always try something. Even paths that lead to dead ends provide insights.
Example 1 (Verifying a Trigonometric Identity)
Verify the identity
sec   1
2
sec 
2
= sin2  .
Solution
Because the left side is more complicated, start with it.
L.S.
=
=
=
sec   1
2
sec 
2
 ta n
2
 1  1
sec 
2
ta n 
2
sec 
2
=
ta n   c o s  
=
s in 
2
2
cos 
2
2
 cos
2

= sin2 
= R.S.
Here is another way to verify the identity in example 1.
L.S.
=
sec   1
=
sec 
2
sec 
2
2
sec 
2

2
1
sec 
2
= 1 – cos 
= sin2 
= R.S.
As you can see, there can be more than one way to verify an identity. Your method may
differ from that used by your instructor or fellow students. Here is a good chance to be creative
and establish your own style, but try to be as efficient as possible.
Example 2 (Combining Fractions Before Using Identities)
Verify the identity
1
1  s in 

=
1
1  s in 
2 sec 
2
Solution
L.S.
=
=
=
=
1
1  s in 

1
1  s in 
1  sin   1  sin 
1 
sin 
 1 
sin  
2
1  s in 
2
2
cos 
2
= 2sec2 
= R.S.
Example 3 (Verifying a Trigonometric Identity)
Verify the identity
 ta n x  1   c o s x  1  =  ta n x .
Solution
By applying identities before multiplying, you obtain the foolowing.
L.S. =  ta n x  1   c o s x  1 
=  s e c x    s in x 
2
2
2
2
2
2
=
2
2

s in x
2
cos x
=   s i n x 
2
 cos x 
=  ta n x
= R.S.
Example 4 (Converting to Sines and Cosines)
2
Verify the identity
tan x + cot x = sec x cosec x
Solution
In this case there appear to be no fractions to add, no products to find, and no opportunities to
use the Pythagorean identities. So, try converting the left side into sines and cosines to see what
happens.
L.S. = tan x + cot x
=
=
=
=
s in x

cos x
cos x
s in x
s in x  c o s x
2
2
c o s x s in x
1
c o s x s in x
1
1
.
c o s x s in x
= sec x cosec x
= R.S.
Recall from algebra that rationalizing the denominator using conjugates is, on occasion,
a powerful simplification technique. A related form of this technique works for simplifying
trigonometric expressions as well. For instance, to simplify
1
1  cos x
, multiply the numerator and
the denominator by 1 + cos x.
1
1  cos x
=
 1  cos x 


1  cos x  1  cos x 
1  cos x
=
1
1  cos x
2
=
1  cos x
2
s in x
=
c o s e c x 1  c o s x
2

This technique is demonstrated in the next example.
Example 5 (Verifying Trigonometric Identities)
Verify the identity
sec y + tan y =
cos y
1  s in y
.
Solution
Begin with the right side. Note that you can create a monomial denominator by multiplying the
numerator and the denominator by 1 + sin y
R.S.
=
cos y
1  s in y
=
 1  s in y 


1  s in y  1  s in y 
=
c o s y  c o s y s in y
=
c o s y  c o s y s in y
cos y
1  s in y
2
2
cos y
=
=
cos y
c o s y s in y

2
2
cos y
1
cos y
s in y

cos y
cos y
= sec y + tan y
= L.S.
In example 1 through 5, you have been verifying trigonometric identities by working with one
side of the equation and converting to the form given on the other side. On occasion, it is
practical to work with each side separately, to obtain one common form equivalent to both
sides. This is illustrated in example 6.
Example 6 (Working with Each Side Separately)
Verify the identity
cot 
2
=
1  cos ec
1  s in 
s in 
Solution
Working with the left side, you have
L.S.
=
cot 
2
1  cos ec
=
cos ec   1
=
 cos ec  1  cos ec  1
2
1  cos ec
1  cos ec
= co s ec   1
Now, simplifying the right side, you have
R.S.
=
=
L.S.
Thus, the identity is verified.
1  s in 
s in 
1
s in 

s in 
s in 
= co s ec   1
= R.S.
In example 7, powers of trigonometric functions are rewritten as more complicated sums of
products of trigonometric functions. This is a common procedure used in calculus.
Example 7 (Three Examples from Calculus)
Verify each identity.
a. tan4 x = tan2 x sec2 x – tan2 x
b. sin3 x cos4 x = (cos4 x – cos6 x)sin x
c. cosec4 x cot x = cosec2 x(cot x + cot3 x)
Solution
a.
L.S.
=
=
=
=
=
tan4 x
(tan2 x)( tan2 x)
tan2 x(sec2 x – 1)
tan2 x sec2 x – tan2 x
R.S.
b.
L.S.
=
=
=
=
=
sin3 x cos4 x
sin2 x cos4 x sin x
(1 – cos2 x) cos4 x sin x
(cos4 x – cos6 x) sin x
R.S.
c.
L.S.
=
=
=
=
=
cosec4 x cot x
cosec2 x cosec2 x cot x
cosec2 x (1+ cot2 x) cot x
cosec2 x (cot x + cot3 x)
R.S.