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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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