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11 SL: Trigonometry Review

Complete the following questions before our next class (all questions are from the red book).

Exercise 7B questions 2ab, 4b
Exercise 7C questions 1de, 3b
Exercise 7D questions 1ef, 9a
Exercise 7E questions 1, 5, 17

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12 HL: Areas and Integrals

Complete the following questions (this was homework from our last class, make sure it’s done for our next class if you haven’t completed it already).

Exercise 22C questions 13, 16cd
Exercise 22D questions 1d, 6, 8
Exercise 22E questions 4, 8, 10, 13

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12 HL: Definite Integrals

Complete as many of the following questions as you can before our next class (we’ll continue to work on these in our next class as well).

Exercise 22A questions 3–8, 12 (just complete the centre column for 3–8 and 12), 14, 16, 21
Exercise 22B questions 1adf, 2ad, 3, 4a

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12 HL: Trigonometric Substitutions

In class we arrived at part of the solution to \(\int \sqrt{2-x^2}\;d x\), using the substitution \(x = \sqrt{2}\sin \theta\). Complete this question by showing that \[\int \sqrt{2-x^2}\;d x=\left( \frac{x}{\sqrt{2}} \right) \sqrt{ 1-\frac{x^2}{2}} +\arcsin \left(\frac{x}{\sqrt{2}}\right) + C\]

Hint
Note that, given our substitution choice, it follows that \(\sin \theta = \frac{x}{\sqrt{2}}\). Can we do something similar for \(\cos \theta\)?

Also complete

Exercise 21E questions 4, 5d, 6
Exercise 21F questions 12ad, 17, 18bd