Aim: Cosine of Two Angles  Course: Alg. 2 & Trig. Aim: How do we find the cosine of the difference and sum of two angles? Do Now: If m  A = 35º and m  B = 20º show which of the following identities is true? cos(A – B) = cos A – cos B cos(A – B) = cosAcosB +sinASinB cos(35º – 20º) = cos 35º – cos 20º cos(15º) = – = = cos 35cos 20 + sin 35sin =

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

Aim: Cosine of Two Angles  Course: Alg. 2 & Trig. 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

Aim: Cosine of Two Angles  Course: Alg. 2 & Trig. 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 sin 2 A + cos 2 A = 1sin 2 B + cos 2 B = 1 (3/5) 2 + cos 2 A = 1 sin 2 B + (5/13) 2 = 1 9/25 + cos 2 A = 1sin 2 B + 25/169 = 1 cos 2 A = 16/25 cos A = 4/5 sin 2 B = 144/169 sin B = 12/13 QIIcos A = -4/5QI

Aim: Cosine of Two Angles  Course: Alg. 2 & Trig. 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)  B: cos B = 5/13, sin B = 12/13  A: sin A = 3/5, cos A = -4/5 cos (A – B) = (-20/65) + (36/65) cos (A – B) = (16/65)

Aim: Cosine of Two Angles  Course: Alg. 2 & Trig. Cosine of Sum of 2 Angles The Cosine of the Sum of 2 Angles: cos (A + B) = cos A cos B – sin A sin B Prove: 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 (-x) = cos (x)sin (-x) = -sin x cos (A + B) = cos A cos B – sin A sinB ex. Show that cos 90 = 0 by using cos( ) cos ( ) = cos 60 cos 30 – sin 60 sin 30 = 0

Aim: Cosine of Two Angles  Course: Alg. 2 & Trig. Model Problem Find the exact value of cos45ºcos15º – sin45ºsin15º cos( ) =cos45ºcos15º – sin45ºsin15º cos(60) = 1/2 The Cosine of the Sum of 2 Angles: cos (A + B) = cos A cos B – sin A sin B Find the exact value of cos 75 by using cos ( ) cos ( ) = cos 45 cos 30 – sin 45 sin 30

Aim: Cosine of Two Angles  Course: Alg. 2 & Trig. 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 sin  = cos(90 –  ) 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 the sine of an angle equals the cosine of its complement, let  = -x

Aim: Cosine of Two Angles  Course: Alg. 2 & Trig. Model Problem

Aim: Cosine of Two Angles  Course: Alg. 2 & Trig. Regents Question

Aim: Cosine of Two Angles  Course: Alg. 2 & Trig. The Product Rule