Math 105 TOPICS IN MATHEMATICS REVIEW OF LECTURES XXXIII April 20 (Mon), 2015 Yasuyuki Kachi. 33. Trigonometry III.
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1 Instructor: Math 105 TOPICS IN MATHEMATICS REVIEW OF LECTURES XXXIII April 20 Mon, 2015 Yasuyuki Kachi Line #: Trigonometry III In the previous lecture we calculated the distance between P π 4, π 4, Q Now, we can more generally consider the following question: π 6, π 6 Question 1 P Find the distance between,, Q φ, φ where and φ are arbitrary [ Solution PQ ] : φ 2 φ φ φ φ φ φ 2 2 φ 2 φ 2 φ 1
2 Now, exactly as we previously did in the case π 4 and φ π 6, we pose: Question 2 R Find the distance between φ, φ, S 1, 0 Explain why PQ and RS are equal Then use that fact and the result of Question 1 above to write up φ in terms of,, φ and φ [ Solution ] : P and Q are both lying in the unit circle such that the angle POQ equals φ R and S are also both lying in the unit circle such that the angle ROS equals φ Hence PQ and RS are naturally equal Now, RS φ 2 1 φ 2 φ 2 2 φ φ φ φ 2 This equals PQ 2 2 φ 2 φ 2
3 So 2 2 φ 2 2 φ 2 φ Solve this for φ : φ φ φ Let s highlight this result: Axiom 1 φ φ φ Now, we shoot for pulling an analogous identity for φ Toward that goal, working on the following two questions Question 1, Question 2 is the correct step to take: Question 1 P Find the distance between,, Q φ, φ where and φ are arbitrary The answer is, P Q 2 2 φ 2 φ The calculation is similar to Question 1 3
4 Question 2 R Find the distance between φ, φ, S 1, 0 Explain why P Q and R S are equal Then use that fact and the result of Question 1 above to write up φ in terms of,, φ and φ [ Solution ] : You can solve this exactly the same way as Question 2: P and Q are both lying in the unit circle such that the angle P OQ equals π φ R and S are also both lying in the unit circle such that the angle 2 R π OS equals 2 φ Hence P Q and R S are naturally equal Now, R S 2 2 φ The calculation is similar to the calculation of RS This equals P Q 2 2 φ 2 φ So 2 2 φ 2 2 φ 2 φ Solve this for φ : φ φ φ 4
5 Let s pair Axiom 1 which we deduced earlier with the above new result, and highlight them together: Axiom 1 φ φ φ Axiom 2 φ φ φ How about analogs for φ and for φ? For that matter, we rely on Axiom 1 and Axiom 2, and proceed as follows: First, set φ α Thus α φ Then the above two identities are φ αφ φ αφ φ αφ α, φ αφ α These two equations are regarded as ax by p, cx dy q, where a φ, b φ, p α, c φ, d φ, q α, x φ, and y φ Clearly ad bc 1 Thus we rely on the result in Supplement : x dp bq, y cp aq 5
6 That is: αφ αφ φ α φ α φ α, φ α At this point, we are free to replace the letter α with Let s go ahead and make that replacement, and highlight: Axiom 3 φ φ φ Axiom 4 φ φ φ In Axiom 1 and Axiom 2, set φ and we get Double angle formula for Double angle formula for version Double angle formula for version Double angle formula for 2 2 6
7 From Axiom 1 and Axiom 3, we have Formula A 2 φ Formula B 2 φ φ φ φ φ, Similarly, from Axiom 2 and Axiom 4, we have Formula C 2 φ φ φ Today s final topic is lim 0 This is somehow very important Don t ask me yet how so As for this, consider P,, P 0 1, 0, T 1, Assume > 0 T is the unique point lying in the extension of the segment OP whose x-coordinate reading is 1 By drawing the figure, we have the area of POP 0 < < the area of circle sector POP 0 the area of TOP 0 7
8 This translates into 2 < 2 < # 2 Here, we use the fact that the area of the entire disc of radius 1 equals π the middle term is 2π times the area of the entire unit disc From we have < 1, whereas from # we have < In short, < < 1 From this it is inferred that lim 0 1 Indeed, lim To highlight: Formula D lim 0 1 8
9 Technically, this last implication is due to the so-called squeeze theorem The squeeze theorem asserts that, if f x < g x < h x, for a < x < b where a is some negative number, a is some positive number, and moreover if lim f x lim h x, x 0 x 0 then these limits further equal lim g x x 0 9
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