€

log [1−(1−sin

2 θ )]

1

(1+ cot2 θ)

1

(sec2 θ − ta

n2 θ −cos2 θ ) =

[sin(θ+θ)]

2

4 sinlog 21

6 θ+1

Chapter Four

SOLUTIONBy «Craig»

€

log[1−(1−sin 2θ )]

1(1+ cot 2θ)

1(sec2θ − tan 2θ −cos2 θ )

=[sin(θ +θ)]2

4 sinlog216θ+1

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.

€

log[1−(1−sin 2θ )]1

(1+ cot2θ)

1(sec2θ − tan 2θ −cos2 θ )

Step 1- Solve the logarithm.

€

log[1−(1−sin 2θ )]1

(1+ cot2θ)

1(sec2θ − tan 2θ −cos2 θ )

€

[1−(1−sin2 θ )]

€

1(1+ cot 2θ)

€

1(sec2θ − tan 2θ −cos2θ )

Step 1- Solve the logartithm.

There are 3 parts to this logarithm:Part A. BasePart B. Base of ArgumentPart C. Exponent of Argument

€

[1− (1− sin2θ)]

€

1(1+ cot2θ)

€

1(sec2θ − tan2θ − cos2θ)

€

[1−(1−sin 2θ )]

Step 1- Solve the logartithm.

Part A. Simplify the Base.

€

[1−(1−sin 2θ )]

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 1Method 2

Method 1 Method 2

€

[1−(1−sin 2θ )]

Step 1- Solve the logartithm. Part A. Simplify the Base

Method 1

€

1− (1− sin2θ)

€

[1−(1−sin 2θ )]

Step 1- Solve the logartithm. Part A. Simplify the Base

Method 1

€

1−1+ sin2θ

Method 1: Expand the original expression.

€

[1−(1−sin 2θ )]

Step 1- Solve the logartithm. Part A. Simplify the Base

Method 1

€

sin2θ

Method 1: Simplify the resulting expression

€

[1−(1−sin 2θ )]

Step 1- Solve the logartithm. Part A. Simplify the Base

Method 2

€

1− (1− sin2θ)

€

[1−(1−sin 2θ )]

Step 1- Solve the logartithm. Part A. Simplify the Base

Method 2

Method 2: Rearrange the identity

€

sin2θ + cos2θ =1 so that

€

cos2θ =1− sin2θ and use it to simplify the expression.

€

1− (cos2θ)

€

[1−(1−sin 2θ )]

Step 1- Solve the logartithm. Part A. Simplify the Base

Method 2

Method 2: Simplify the resulting expression

€

sin2θ

Step 1- Solve the logartithm. Part A. Simplify the Base

Therefore, the Base is equal to

€

sin2θ

€

sin2θ

€

[1− (1− sin2θ)]

€

=

€

[1−(1−sin 2θ )]

Step 1- Solve the logartithm.

Part B. Simplify the Base of Argument.

€

1(1+ cot 2θ)

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

€

1(1+ cot 2θ)

There are two methods that can be used to get the simplified version of this expression:

Method 1Method 2

Method 1 Method 2

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

€

1(1+ cot 2θ)Method 1

€

1(1+ cot2θ)

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

€

1(1+ cot 2θ)Method 1

€

1sin2θsin2θ

+cos2θsin2θ

Method 1: Recognize that

€

cot 2θ is the same thing as

€

cos2θsin2θ and

expand it as such. Also, rewrite “1” so it has the same LCD.

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

€

1(1+ cot 2θ)Method 1

€

1sin2θ + cos2θ

sin2θ

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

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

€

1(1+ cot 2θ)Method 1

€

11

sin2θ

Method 1: Refer to the Pythagorean identity (

€

sin2θ + cos2θ =1) and use it to simplify the numerator of the fraction in the denominator.

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

€

1(1+ cot 2θ)Method 1

€

sin2θ

Method 1: Multiply “1” by the reciprocal of the fraction in the

denominator (

€

sin2 θ ).

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

€

1(1+ cot 2θ)Method 2

€

1(1+ cot2θ)

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

€

1(1+ cot 2θ)Method 2

€

1csc2θ

Method 2: Recognize that the Pythagorean Identity

€

csc2θ − cot2θ =1 applies to the denominator and simplify it.

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

€

1(1+ cot 2θ)

€

11

sin2θMethod 2: Recognize that

€

csc2θ is eqivalent to the reciprocal of

€

sin2θ (

€

1sin2θ ) and rewrite it as such.

Method 2

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

€

1(1+ cot 2θ)Method 2

€

sin2θ

Method 2: Multiply “1” by the reciprocal of the fraction in the

denominator (

€

sin2 θ ).

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

€

1(1+ cot 2θ)

€

sin2θ

Therefore, the Base of the Argument

is equal to

€

sin2θ .

€

1(1+ cot2θ)

€

=

Step 1- Solve the logartithm.

So far the logarithm goes from this...€

log[1−(1−sin 2θ )]1

(1+ cot2θ)

1(sec2θ − tan 2θ −cos2 θ )

€

[1− (1− sin2θ)]

€

1(1+ cot2θ)

€

1(sec2θ − tan2θ − cos2θ)

€

log[1−(1−sin 2θ )]1

(1+ cot2θ)

1(sec2θ − tan 2θ −cos2 θ )

€

1(sec2θ − tan2θ − cos2θ)

Step 1- Solve the logartithm.

...to this.€

sin2θ

€

sin2θ

€

log[1−(1−sin 2θ )]1

(1+ cot2θ)

1(sec2θ − tan 2θ −cos2 θ )

€

1(sec2θ − tan2θ − cos2θ)

€

sin2θ

€

sin2θ

Step 1- Solve the logartithm.

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.

€

log[1−(1−sin 2θ )]1

(1+ cot2θ)

1(sec2θ − tan 2θ −cos2θ )

=1

(sec2θ − tan2θ − cos2θ)

Step 1- Solve the logartithm.

Therefore, the logarithm is equal to

€

1(sec2θ − tan2θ − cos2θ) .

€

log[1−(1−sin 2θ )]

1(1+ cot 2θ)

1(sec2θ − tan 2θ −cos2 θ )

=[sin(θ +θ)]2

4 sinlog216θ+1

Step 1- Solve the logartithm.

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

Step 1- Solve the logartithm.

...it looks like this

€

1(sec2θ − tan2θ − cos2θ)

=[sin(θ + θ)]2

4 sinlog2 16θ+1

Step 2- Prove the identity.

€

1(sec2θ − tan2θ − cos2θ)

=[sin(θ + θ)]2

4sinlog2 16θ+1

€

1(sec2θ − tan2θ − cos2θ)

=[sin(θ + θ)]2

4 sinlog2 16θ+1

Step 2- Prove the identity.

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

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity.

Side 1.

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1There are two methods that can be used to get the simplified version of this expression:

Method 1Method 2

Method 1 Method 2

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

Method 1

€

1(sec2θ − tan2θ − cos2θ)

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

Method 1

€

1

[( 1cos2θ

) − ( sin2θ

cos2θ)− cos2θ]

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

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

Method 1

€

1

[(1− sin2θ

cos2θ)− cos2θ]

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

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

Method 1

€

1

[(cos2θ

cos2θ)− cos2θ]

Method 1: Due to the Pythagorean identity (

€

sin2θ + cos2θ =1), the fraction in the denominator is simplified to

€

cos2θcos2θ

.

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

Method 1

€

1(1− cos2θ)

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

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

Method 1

€

1sin2θ

Method 1: Due to the Pythagorean identity (

€

sin2θ + cos2θ =1), the denominator is simplified to

€

sin2θ .

Method 1:

€

1sin2θ can also be rewritten as

€

csc2θ

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

Method 1

€

csc2θ

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

€

1(sec2θ − tan2θ − cos2θ)

Method 2

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

€

1(1− cos2θ)

Method 2

Method 2: Recognize that the the Pythagorean Identity

€

sec2 θ − tan2 θ =1 applies to the denominator of the expression and simplify it as such.

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

€

1sin2θ

Method 2

Method 2: Using the Pythagorean identity (

€

sin2θ + cos2θ =1), simplify the denominator to

€

sin2θ .

Method 2:

€

1sin2θ can also be rewritten as

€

csc2θ

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity. Side 1

€

csc2θ

€

1(sec2θ − tan2θ − cos2θ)

=[sin(θ + θ)]2

4 sinlog2 16θ+1

Step 2- Prove the identity.

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

€

csc2θ =[sin(θ + θ)]2

4 sinlog216θ+1

Step 2- Prove the identity.

...it looks like this

€

[sin(θ + θ)]2

4sinlog2 16θ+1

Step 2- Prove the identity.

Side 2.

Step 2- Prove the identity. Side 2There are two methods that can be used to get the simplified version of this expression:

Method 1Method 2

Method 1 Method 2

€

[sin(θ +θ)]2

4 sinlog216θ+1

Step 2- Prove the identity. Side 2

Method 1

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

€

[sin(θ +θ)]2

4sinlog2 16θ+1

Step 2- Prove the identity. Side 2

Method 1

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

€

[sin(θ +θ)]2

4 sin4 θ+1

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.

€

2x =16).

Step 2- Prove the identity. Side 2

Method 1

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

€

(2sinθ cosθ)2

4 sin4 θ+1

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)

Step 2- Prove the identity. Side 2

Method 1

€

[sin(θ +θ)]2

4 sinlog216θ+1

€

(2)(sinθ)(cosθ)(2)(sinθ)(cosθ)(2)(2)(sinθ)(sinθ)(sinθ)(sinθ)

+1

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

Step 2- Prove the identity. Side 2

Method 1

€

[sin(θ +θ)]2

4 sinlog216θ+1

€

(2)(sinθ)(cosθ)(2)(sinθ)(cosθ)(2)(2)(sinθ)(sinθ)(sinθ)(sinθ)

+1

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

Step 2- Prove the identity. Side 2

Method 1

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

Method 1: This is the resulting expression.

€

(cosθ)2

(sinθ)2+1

Step 2- Prove the identity. Side 2

Method 1

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

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

€

cos2θsin2θ

+sin2θsin2θ

Step 2- Prove the identity. Side 2

Method 1

€

[sin(θ +θ)]2

4 sinlog216θ+1

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

€

cos2θ + sin2θsin2θ

Step 2- Prove the identity. Side 2

Method 1

€

[sin(θ +θ)]2

4 sinlog216θ+1

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

€

1sin2θ

Step 2- Prove the identity. Side 2

Method 1

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

Method 1: Fianlly,

€

1sin2θ can also be rewritten as

€

csc2θ

€

csc2θ

Step 2- Prove the identity. Side 2

Method 2

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

€

[sin(θ +θ)]2

4sinlog2 16θ+1

Step 2- Prove the identity. Side 2

Method 2

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

€

[sin(θ + θ)]2

4sin4 θ+1

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

Step 2- Prove the identity. Side 2

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

€

(2sinθ cosθ)2

4sin4 θ+1

Method 2

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

Step 2- Prove the identity. Side 2

€

[sin(θ +θ)]2

4 sinlog216θ+1

€

(2)(sinθ)(cosθ)(2)(sinθ)(cosθ)(2)(2)(sinθ)(sinθ)(sinθ)(sinθ)

+1

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

Method 2

€

(cosθ)2

(sinθ)2+1

Step 2- Prove the identity. Side 2

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

Method 2: ...THIS!

Method 2

€

(cosθ)2

(sinθ)2+1

Step 2- Prove the identity. Side 2

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

Method 2

Method 2: However, instead of rewriting “1” as a fraction, due to the fact that

€

(cosθ) 2

(sinθ) 2 is an identity itself,

rewrite the expression as...

Step 2- Prove the identity. Side 2

€

[sin(θ + θ)]2

4 sinlog2 16θ+1

Method 2

€

cot2θ +1

Step 2- Prove the identity. Side 2

€

[sin(θ +θ)]2

4 sinlog216θ+1

Method 2

Method 2: Now, apply the Pythagorean Identity

€

csc2θ − cot 2θ =1 and rewrite the expression as

€

csc2θ .

€

csc2θ

Step 2- Prove the identity.

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

[sin(θ +θ)]2

4 sinlog216θ+1

€

csc2θ

€

=

Step 2- Prove the identity.

...SINCE...

€

1(sec2θ − tan2θ − cos2θ)

€

csc2θ

€

=

Step 2- Prove the identity.

...AND...

€

csc2θ

€

=

€

[sin(θ +θ)]2

4sinlog2 16θ+1

Step 2- Prove the identity.

...THEN...

€

=

€

[sin(θ +θ)]2

4sinlog2 16θ+1

€

1(sec2θ − tan2θ − cos2θ)

Step 2- Prove the identity.

Q

Step 2- Prove the identity.

E

Step 2- Prove the identity.

D

€

log[1−(1−sin 2θ )]

1(1+ cot 2θ)

1(sec2θ − tan 2θ −cos2 θ )

=[sin(θ +θ)]2

4 sinlog216θ+1

€

csc2θ = csc2θQ.E.D.

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