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Aim: How do we find the cosine of the difference and sum of two angles?

Aim: How do we find the cosine of the difference and sum of two angles?. If m A = 35º and m B = 20º show which of the following identities is true?. Do Now:. cos(A – B) = cos A – cos B. cos(35º – 20º) = cos 35º – cos 20º. cos(15º) = .8191520443 – .9396926208. .9659258263 = -.1205405765.

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Aim: How do we find the cosine of the difference and sum of two angles?

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  1. Aim: How do we find the cosine of the difference and sum of two angles? If mA = 35º and mB = 20º show which of the following identities is true? Do Now: cos(A – B) = cos A – cos B cos(35º – 20º) = cos 35º – cos 20º cos(15º) = .8191520443 – .9396926208 .9659258263 = -.1205405765 cos(A – B) = cosAcosB +sinASinB .9659258263 = cos 35•cos 20 + sin 35•sin20 .9659258263 = .9659258263

  2. P(cos B, sin B) Q(cos A, sin A) (x,y) A – B B A The Cosine of the Difference of 2 Angles cos (A – B) = cos A cos B + sin A sin B Outline of Proof – Cos(A – B) y 1 O x -1 1 Find the length of PQ using both the Law of Cosines and the distance formula, equating the two to arrive at:

  3. Model Problem Use the identity for the cosine of the difference of two angle measures to prove that cos (180º – x) = -cos x. cos (A – B) = cos A cos B + sin A sin B Substitute 180 for A and x for B: cos (180 – x) = cos 180 cos x + sin 180 sin x simplify: cos (180 – x) = -1 • cos x + 0 • sin x cos (180 – x) = -cos x

  4. Model Problem If sin A = 3/5 and A is in QII, and cos B = 5/13 and B is in QI, find cos (A – B). cos (A – B) = cos A cos B + sin A sin B to use this you need to know both sine and cosine values for both A and B. HOW? Pythagorean Identity sin2 A + cos2 A = 1 sin2 B + cos2 B = 1 sin2 B + (5/13)2 = 1 (3/5)2 + cos2 A = 1 9/25 + cos2 A = 1 sin2 B + 25/169 = 1 sin2 B = 144/169 cos2 A = 16/25 sin B = 12/13 cosA = 4/5 QII cosA = -4/5 QI

  5. A: sin A = 3/5, cosA = -4/5 B: cos B = 5/13, sin B = 12/13 Model Problem (con’t) If sin A = 3/5 and A is in QII, and cos B = 5/13 and B is in QI, find cos (A – B). cos (A – B) = cos A cos B + sin A sin B Substitute & simplify: cos (A – B) = (-4/5)(5/13) + (3/5)(12/13) cos (A – B) = (-20/65) + (36/65) cos (A – B) = (16/65)

  6. cos (-x) = cos (x) sin (-x) = -sin x The Cosine of the Sum of 2 Angles: cos (A + B) = cos A cos B – sin A sin B Cosine of Sum of 2 Angles Prove: cos (A + B) = cos A cos B – sin A sin B cos (A + B) = cos (A - (-B)) cos (A + B) = cos A cos(-B) + sin A sin(-B) cos (A + B) = cos A(cos B) + sin A(-sinB) cos (A + B) = cos A cos B – sin A sinB ex. Show that cos 90 = 0 by using cos(60 + 30) cos (60 + 30) = cos 60 cos 30 – sin 60 sin 30 = 0

  7. The Cosine of the Sum of 2 Angles: cos (A + B) = cos A cos B – sin A sin B Model Problem Find the exact value of cos45ºcos15º – sin45ºsin15º cos(45 + 15) =cos45ºcos15º – sin45ºsin15º cos(60) = 1/2 Find the exact value of cos 75 by using cos (45 + 30) cos (45 + 30) = cos 45 cos 30 – sin 45 sin 30

  8. More Proofs Prove: cos (-x) = cos x cos (A – B) = cos A cos B + sin A sin B cos (0 – x) = cos 0 cos (x) + sin 0 sin (x) cos (0 – x) = 1 • cos (-x) + 0 • sin (-x) cos (-x) = cos (x) Prove: sin (-x) = -sin x the sine of an angle equals the cosine of its complement sin  = cos(90 – ) , let  = -x sin (-x) = cos(90 – (-x)) = = cos (90 + x) = cos (x + 90) = cos (x – (-90) = cos x cos (-90) + sin x sin (-90) = cos x • 0 + sin x • -1 = -sin x

  9. Model Problem

  10. Regents Question

  11. The Product Rule

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