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M1410504and05

# M1410504and05 - (Sections 5.4 and 5.5 More Trig Identities...

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(Sections 5.4 and 5.5: More Trig Identities) 5.42 SECTIONS 5.4 and 5.5: MORE TRIG IDENTITIES PART A: A GUIDE TO THE HANDOUT See the Handout on my website. The identities (IDs) may be derived according to this flowchart: In Calculus: The Double-Angle and Power-Reducing IDs are most commonly used among these, though we will discuss a critical application of the Sum IDs in Part C . Some proofs are on pp.403-5 . See p.381 for notes on Hipparchus, the “inventor” of trig, and the father of the Sum and Difference IDs.

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(Sections 5.4 and 5.5: More Trig Identities) 5.43 PART B: EXAMPLES Example: Finding Trig Values Find the exact value of sin15 . Note: Larson uses radians to solve this in Example 2 on p.381 , but degrees are usually easier to deal with when applying these identities, since we don’t have to worry about common denominators. Solution (Method 1: Difference ID) We know trig values for 45 and 30 , so a Difference ID should work. sin15 = sin 45 30 ( ) Use: sin u v ( ) = sin u cos v cos u sin v = sin45 cos30 cos45 sin30 = 2 2 3 2 2 2 1 2 = 6 4 2 4 = 6 2 4 Warning: 6 2 4 . We do not have sum and difference rules for radicals the same way we have product and quotient rules for them.
(Sections 5.4 and 5.5: More Trig Identities) 5.44 Solution (Method 2: Half-Angle ID) We know trig values for 30 , so a Half-Angle ID should work. sin15 = sin 30 2 Use: sin θ 2 = ± 1 cos θ 2 = ± 1 cos30 2 = ± 1 3 2 2 2 2 = ± 2 3 4 = ± 2 3 2 We know sin15 > 0 , since 15 is an acute Quadrant I angle. We take the “+” sign. = 2 3 2 In fact, 2 3 2 is equivalent to 6 2 4 , our result from Method 1.

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