Chapter 4

2
)] + 1
+θ 6
n(θ log 21 θ
[si
= 4 sin
Chapter Four 2 θ)
1
cos
2 θ−
− tan
2θ
(sec
1
SOLUTIONθ)
2
ot
+c
1
(
θ )]
2
− sin
log 1−(1
[ By «Craig»

1
2
1 (sec θ −tan θ −cos θ ) [sin(θ + θ)]
2 2 2

log[1−(1−sin 2 θ )] = +1
2 log 2 16
(1+ cot θ) 4 sin θ

At first glance, this problem can look a little bit complicated
and intimidating. A good way to make it easier to solve is to
break it down:
Step 1- Solve the logarithm.
Step 2- Prove the resulting identity.

1
1 (sec 2 θ −tan 2 θ −cos2 θ )
log[1−(1−sin 2 θ )] 2
(1+ cot θ)

Step 1- Solve the logarithm.

1
1
2 2 2
(sec θ − tan θ − cos θ)
log[1−(1−sin 2 θθ)] 2
2
[1− (1− sin )]
(1+ cot θ)
(1+
€
€
€
Step 1- Solve the logartithm.
€
There are 3 parts to this logarithm:
€
Part A. Base
Part B. Base of Argument
Part C. Exponent of Argument

2
[1−(1−sin θ )]

Part A. Simplify the Base.

2
[1−(1−sin θ )]

Method 2
Method 1

Step 1- Solve the logartithm. Part A. Simplify the Base
There are two methods that can be used to get the simplified
version of this expression:
Method 1
Method 2

2
[1−(1−sin θ )]
Method 1

2
1− (1− sin θ)


€

2
[1−(1−sin θ )]
Method 1

2
1−1+ sin θ


Method 1: Expand the original expression.

€

2
[1−(1−sin θ )]
Method 1

2
sin θ

Method 1: Simplify the resulting expression

2
[1−(1−sin θ )]
Method 2

2
1− (1− sin θ)


€

2
[1−(1−sin θ )]
Method 2

2
1− (cos θ)


sin2 θ + cos2 θ =1 so that
Method 2: Rearrange the identity
cos 2 θ = 1− sin2 θ and use it to simplify the expression.

€ €

2
[1−(1−sin θ )]
Method 2

2
sin θ

Method 2: Simplify the resulting expression

2
[1−(1−sin θ )]

2
2
[1− (1− sin θ)] = sin θ


2
sin θ
Therefore, the Base is equal to

€ €

1
2
(1+ cot θ)

Part B. Simplify the Base of Argument.

1
2
(1+ cot θ)

Method 2
Method 1
€
Step 1- Solve the logartithm. Part B. Simplify the Base of Arugement
Method 1
Method 2

1
2
(1+ cot θ)
Method 1
1
2
(1+ cot θ)
€

1
2
(1+ cot θ)
Method 1
1
2 2
sin θ cos θ
+
€ 2 2
sin θ sin θ

cos2 θ
2
Method 1: Recognize that cot θ is the same thing as and sin2 θ
expand it as such. Also, rewrite “1” so it has the same LCD.
€

€ €

1
2
(1+ cot θ)
Method 1
1
2 2
sin θ + cos θ
€ 2
sin θ

Method 1: Add the two fractions together to get one fraction.

€

1
2
(1+ cot θ)
Method 1
1
1
€ 2
sin θ

sin2 θ + cos2 θ =1)
Method 1: Refer to the Pythagorean identity (
and use it to simplify the numerator of the
fraction in the denominator.

€
€

1
2
(1+ cot θ)
Method 1

2
sin θ
€

Method 1: Multiply “1” by the reciprocal of the fraction in the
2
denominator ( sin θ ).

€

1
2
(1+ cot θ)
Method 2
1
2
(1+ cot θ)
€

1
2
(1+ cot θ)
Method 2
1
2
csc θ
€

Method 2: Recognize that the Pythagorean Identity
csc2 θ − cot 2 θ = 1 applies to the denominator and
simplify it.

€

1
2
(1+ cot θ)
Method 2
1
1
€ 2
sin θ

csc2 θ is eqivalent to the reciprocal
Method 2: Recognize that
2
of sin θ ( sin θ ) and rewrite it as such.
1
2

€

€
€€

1
2
(1+ cot θ)
Method 2

2
sin θ
€

Method 2: Multiply “1” by the reciprocal of the fraction in the
2
denominator ( sin θ ).

€

1
2
(1+ cot θ)

1 2
sin θ
=
2
€(1+ cot θ)

Therefore, the Base of the Argument
2
sin θ .
is equal to
€

1
1
2 2 2
log[1−(1−sin 2 θθ)] 2
2
[1− (1− sin )]
(1+ cot θ)
(1+
€
€

€
So far the logarithm goes from this...

1
1
12 (sec 2 θθ−tan 2 θθ−cos2 θθ))
2 2 2
(sec − tan − cos
log[1−(1−sinθ2 θ )] sin θ
sin2 2
(1+ cot θ)
€
€

€
...to this.

1
1
12 (sec 2 θθ−tan 2 θθ−cos2 θθ))
2 2 2
(sec − tan − cos
log[1−(1−sinθ2 θ )] sin θ
sin2 2
(1+ cot θ)
€
€

€
We can recognize this about the logarithm: The Base is the
same as the Base of the Argument. If this is the case, no
matter what the Exponent of the Argument is, the entire
logarithm is equal to the Exponent of the Argument.

1
1 (sec2 θ −tan 2 θ −cos2 θ ) 1
log[1−(1−sin 2 θ )] =2
2 2 2
(1+ cot θ) (sec θ −tan θ − cos θ)


1
Therefore, the logarithm is equal to (sec 2 θ −tan 2 θ − cos2 θ) .

€

1
2
2 2 2

log[1−(1−sin 2 θ )] = +1
2 log 2 16


Now, we can take a look at the original identity.
Since we have solved the logarithm the identity no
longer looks like this...

2
1 [sin(θ + θ)]
+1
=
2 2 2 log 2 16
(sec θ − tan θ − cos θ) 4 sin θ

€ Step 1- Solve the logartithm.

...it looks like this

2
1 [sin(θ + θ)]
= +1
2 2 2 log 2 16
(sec θ − tan θ − cos θ) 4sin θ

€
Step 2- Prove the identity.

2
1 [sin(θ + θ)]
+1
=
2 2 2 log 2 16

€ Step 2- Prove the identity.

To prove the identity, solve the two sides separately:
Side 1
Side 2

1
2 2 2

Side 1.

1
2 2 2

Method 2
Method 1
€

Step 2- Prove the identity. Side 1
Method 1
Method 2

1
2 2 2
Method 1

1
€ 2 2 2

€

1
2 2 2
Method 1
1
2
1 sin θ
€ 2
[( 2 ) − ( 2 ) − cos θ]
cos θ cos θ

Method 1: Simplify the first two terms so that they are
expressed in terms of sine and/or cosine.
€

1
2 2 2
Method 1
1
2
€ 1− sin θ 2
[( ) − cos θ]
2
cos θ

Method 1: Subtract the two fractions in the denominator to
get one fraction.

€

1
2 2 2
Method 1
1
2
€ cos θ 2
[( 2 ) − cos θ]
cos θ

2 2
Method 1: Due to the Pythagorean identity ( sin θ + cos θ =1),
2
cos θ
the fraction in the denominator is simplified to cos θ 2

€
.
€ €

1
2 2 2
Method 1

1
€ 2
(1− cos θ)

Method 1: Simplify the fraction in the denominator to “1”.

€

1
2 2 2
Method 1

1
€
2
sin θ

2 2
Method 1: Due to the Pythagorean identity ( sin θ + cos θ =1),
2
the denominator is simplified to sin θ .

€ €
€

1
2 2 2
Method 1

2
csc θ
€


csc 2 θ
1
Method 1: can also be rewritten as
sin2 θ

€

€

1
2 2 2
Method 2

1
€ 2 2 2

€

1
2 2 2
Method 2

1
€
2
(1− cos θ)

Method 2: Recognize that the the Pythagorean Identity
sec 2 θ −tan 2 θ =1 applies to the denominator of the
expression and simplify it as such.

€

1
2 2 2
Method 2

1
€
2
sin θ

2
θ + cos2 θ =1),
Method 2: Using the Pythagorean identity ( sin
2
simplify the denominator to sin θ .

€
€

1
2 2 2

2
csc θ
€


csc 2 θ
1
Method 2: can also be rewritten as
sin2 θ

€

€

2
1 [sin(θ + θ)]
+1
=
2 2 2 log 2 16

€ Step 2- Prove the identity.

So far the we have solved Side 1 of the identity. Now, it no
longer looks like this...

2
[sin(θ + θ)]
2
csc θ = +1
log 2 16
4 sin θ

€

...it looks like this

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ

Side 2.

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ

Method 2
Method 1
€

Method 1
Method 2

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 1
2
[sin(θ + θ)]
+1
€ log 2 16
4 sin θ

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 1
2
[sin(θ + θ)]
+1
€ 4
4 sin θ

Method 1: Simplify the logarithm that is an exponent for
one of the terms in the denominator. This can be
done by converting it into a power (ie. 2 x =16 ).

€

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 1
2
(2sinθ cosθ)
+1
€ 4
4 sin θ

Method 1: The expression inside the brackets can be
recognized as one of the “Double Angle Identity”
and therefore can be simplified to 2sinθ cosθ .(see last slide)

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 1
(2)(sinθ)(cosθ)(2)(sinθ)(cosθ)
+1
€ (2)(2)(sinθ)(sinθ)(sinθ)(sinθ)


Method 1: Now both the numerator and the denominator
can be expanded. (Expand as much as possible).

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 1
+1


Method 1: Now we can reduce many parts of the expression
and simplify the remaining terms.

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 1
2
(cosθ)
+1
2
€ (sinθ)

Method 1: This is the resulting expression.

€

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 1
2 2
cos θ sin θ
+2
2
sin θ sin θ
€


Method 1: From there we can rewrite “1” as a fraction with
the same denominator as the resulting fraction.

€

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 1
2 2
cos θ + sin θ
2
sin θ
€


Method 1: After that, it simply becomes a matter of adding
the two fractions together...

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 1
1
2
€
sin θ

Method 1: ...and applying the Pythagorean Identity.

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 1

2
csc θ
€


csc 2 θ
1
Method 1: Fianlly, can also be rewritten as
sin2 θ

€

€

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 2
2
[sin(θ + θ)]
+1
€ log 2 16
4 sin θ

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 2
2
[sin(θ + θ)]
+1
€ 4
4 sin θ

Method 2: The second method starts out the same as the
first, solving the logarithm in the exponent of the
denominator.

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 2
2
(2sinθ cosθ)
+1
€ 4
4sin θ

Method 2: Also the same as the first method, we recognize
the expression inside the brackets as a “Double
Angle Identity”...

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 2
+1


Method 2: The expression, once again, is then expanded;
like terms are reduced; and remaining terms are
simplified to get...

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 2
2
(cosθ)
+1
2
€ (sinθ)

...THIS!
Method 2:

€

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 2
2
(cosθ)
+1
2
€ (sinθ)

Method 2: However, instead of rewriting “1” as a fraction,
due to the fact that (cosθ) is an identity itself,
2

2
(sinθ)

rewrite the expression as...
€

€

2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 2

2
cot θ +1
€


2
[sin(θ + θ)]
+1
log 2 16
4 sin θ
Method 2

2
csc θ
€


2 2
Method 2: Now, apply the Pythagorean Identity csc θ − cot θ =1
csc 2 θ .
and rewrite the expression as

€

€

2
2
[sin(θ + θ)]
csc θ
=
+1
log 2 16
4 sin θ


Now that we have solved Side 2, we can safely say...

€


...SINCE...

1 2
= csc θ
2 2 2


...AND...

2
[sin(θ + θ)] 2
= csc θ
+1
log 2 16
4sin θ


...THEN...

2
1 [sin(θ + θ)]
= +1
2 2 2
(sec θ − tan θ − cos θ) log 2 16
4sin θ

€


Q


E


D

1
2
2 2 2

log[1−(1−sin 2 θ )] = +1
2 log 2 16

2 2
csc θ = csc θ
Q.E.D.

http://www.slideshare.net/dkuropatwa/precal-40s-trigonometric-identities-math-dictionary/1

Chapter 4

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Chapter 4