Chapter 4

log [1−(1−sin

2 θ )]

(1+ cot2 θ)

(sec2 θ − ta

n2 θ −cos2 θ ) =

[sin(θ+θ)]

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)

Step 1- Solve the logarithm.

log[1−(1−sin 2θ )]1

(1+ cot2θ)

[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 )]

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 )]

Method 1

1− (1− sin2θ)

[1−(1−sin 2θ )]

Method 1

1−1+ sin2θ

Method 1: Expand the original expression.

[1−(1−sin 2θ )]

Method 1

sin2θ

Method 1: Simplify the resulting expression

[1−(1−sin 2θ )]

Method 2

1− (1− sin2θ)

[1−(1−sin 2θ )]

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 )]

Method 2

Method 2: Simplify the resulting expression

sin2θ

Therefore, the Base is equal to

sin2θ

[1− (1− sin2θ)]

[1−(1−sin 2θ )]

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

1(1+ cot 2θ)Method 1

1(1+ cot2θ)

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.

1sin2θ + cos2θ

sin2θ

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

sin2θ

Method 1: Refer to the Pythagorean identity (

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

sin2θ

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

denominator (

sin2 θ ).

1(1+ cot2θ)

1csc2θ

Method 2: Recognize that the Pythagorean Identity

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

1(1+ cot 2θ)

sin2θMethod 2: Recognize that

csc2θ is eqivalent to the reciprocal of

sin2θ (

1sin2θ ) and rewrite it as such.

Method 2

sin2θ

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

denominator (

sin2 θ ).

1(1+ cot 2θ)

sin2θ

Therefore, the Base of the Argument

is equal to

sin2θ .

1(1+ cot2θ)

So far the logarithm goes from this...€

log[1−(1−sin 2θ )]1

(1+ cot2θ)

[1− (1− sin2θ)]

1(1+ cot2θ)

log[1−(1−sin 2θ )]1

(1+ cot2θ)

...to this.€

sin2θ

log[1−(1−sin 2θ )]1

(1+ cot2θ)

sin2θ

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 )

(sec2θ − tan2θ − cos2θ)

Therefore, the logarithm is equal to

1(sec2θ − tan2θ − cos2θ) .

log[1−(1−sin 2θ )]

1(1+ cot 2θ)

=[sin(θ +θ)]2

4 sinlog216θ+1

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

...it looks like this

=[sin(θ + θ)]2

4 sinlog2 16θ+1

Step 2- Prove the identity.

=[sin(θ + θ)]2

4sinlog2 16θ+1

=[sin(θ + θ)]2

4 sinlog2 16θ+1

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

Side 1.

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

Step 2- Prove the identity. Side 1

Method 1

[( 1cos2θ

) − ( sin2θ

cos2θ)− cos2θ]

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

Method 1

[(1− sin2θ

cos2θ)− cos2θ]

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

Method 1

[(cos2θ

cos2θ)− cos2θ]

Method 1: Due to the Pythagorean identity (

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

cos2θcos2θ

Method 1

1(1− cos2θ)

Method 1: Simplify the fraction in the denominator to “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

Method 1

csc2θ

Method 2

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.

1sin2θ

Method 2

Method 2: Using the Pythagorean identity (

sin2θ + cos2θ =1), simplify the denominator to

sin2θ .

Method 2:

csc2θ

=[sin(θ + θ)]2

4 sinlog2 16θ+1

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

...it looks like this

[sin(θ + θ)]2

4sinlog2 16θ+1

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

Method 1

[sin(θ + θ)]2

4 sinlog2 16θ+1

[sin(θ +θ)]2

4sinlog2 16θ+1

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).

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)

Method 1

[sin(θ +θ)]2

4 sinlog216θ+1

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

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

Method 1

[sin(θ +θ)]2

4 sinlog216θ+1

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

Method 1

[sin(θ + θ)]2

4 sinlog2 16θ+1

Method 1: This is the resulting expression.

(cosθ)2

(sinθ)2+1

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

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

Method 1

[sin(θ +θ)]2

4 sinlog216θ+1

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

1sin2θ

Method 1

[sin(θ + θ)]2

4 sinlog2 16θ+1

Method 1: Fianlly,

csc2θ

Method 2

[sin(θ + θ)]2

4 sinlog2 16θ+1

[sin(θ +θ)]2

4sinlog2 16θ+1

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.

[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”...

[sin(θ +θ)]2

4 sinlog216θ+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

[sin(θ + θ)]2

4 sinlog2 16θ+1

Method 2: ...THIS!

Method 2

(cosθ)2

(sinθ)2+1

[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...

[sin(θ + θ)]2

4 sinlog2 16θ+1

Method 2

cot2θ +1

[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

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

[sin(θ +θ)]2

4 sinlog216θ+1

csc2θ

...SINCE...

csc2θ

...AND...

csc2θ

[sin(θ +θ)]2

4sinlog2 16θ+1

...THEN...

[sin(θ +θ)]2

4sinlog2 16θ+1

log[1−(1−sin 2θ )]

1(1+ cot 2θ)

=[sin(θ +θ)]2

4 sinlog216θ+1

csc2θ = csc2θQ.E.D.

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

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