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2. Prove the following Identity using two different methods:

2. Prove the following Identity using two different methods:

²sin

²tan

For this question. You can work with the more complicated side. In this case it is the left side.

Method # 1:

Line of Separation / “Great Wall of China”

1}

1} In step one, we realized that we can multiply by reciprocal of the bottom fraction instead of dividing.

2}

2} For step two you can think of tanθ as sinθ/cosθ.

3} In step three we can multiply sinθ by sinθ/cosθ.3}

4}

4} Here we can multiply sin²θ/cosθ out by 1/cosθcos³θ. We can also factor out cos²θ in the denominator.

5}

5} In this last step we can say that sin²θ/cos²θ is tan²θ and by the Pythagorean identity (1-cos²θ) is sin²θ.

QED

sin²θ

tan²θ

cos²-1cos²θ

sin²θθcoscos²θ

sin²θ

cos³θcosθ

1

cosθ

sin²θ

cos³θcosθcosθsinθ

sinθ

cos³cos

tanθsinθ

tanθcos³θcosθ

sinθ

4

²sin

²tan

QED

²sin

²tan

Method # 2:

Line of Separation / “Great Wall of China”

Working with both sides can also work:

1} Here we can say that tan²θ is equal to sin²θ/cos²θ.~by saying that, it can be simplified to 1/cos²θ.

1}

²cos

1²sin²cos²sin²sin

²tan[Right Side]

2}

2} Here we recognize cosθ-cos³θ as cosθ - (cosθcos²θ) or cosθ - cosθ (1-sin²θ). Also, tanθ as sinθ/cosθ.

[Left Side]

3}

3} In this step we can multiply cosθ by (1-sin²θ).

4}

4} By multiplying by the reciprocal of sinθ/cosθ we can see that when cosθ gets multiplied out, we get cos²θ-cos²θ+sin²θcosθ. The two cos²θ’s cancel and you’re left with sin²θcosθ.

5}

5} In this last step we see that sin²θcosθ/sinθ leaves us sinθcosθ. The sinθ’s reduce leaving us 1/cosθ.

QED

cos

1cossin

sinsincos²sin

sincossin

²sincoscossincossin

)²sin1(coscossintan

³coscossin

²cos

1²sin²cos²sin²sin

²tan

QED

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