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 mA = 35º and mB = 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) = .8191520443 – .9396926208 .9659258263 = -.1205405765 .9659258263 = cos 35•cos 20 + sin 35•si .9659258263 = .9659258263
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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.
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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 mA = 35º and mB = 20º show whichof 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) = .8191520443 – .9396926208
.9659258263 = -.1205405765
.9659258263 = cos 35•cos 20 + sin 35•sin20
.9659258263 = .9659258263
Aim: Cosine of Two Angles Course: Alg. 2 & Trig.
Outline of Proof – Cos(A – B)y
x1-1
1
AB
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 differenceof 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 andcosine values for both A and B. HOW?
Pythagorean Identitysin2 A + cos2 A = 1 sin2 B + cos2 B = 1
(3/5)2 + cos2 A = 1 sin2 B + (5/13)2 = 1
9/25 + cos2 A = 1 sin2 B + 25/169 = 1
cos2 A = 16/25cos A = 4/5
sin2 B = 144/169sin B = 12/13
QII cos A = -4/5 QI
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
cos (A + B) = cos A cos B – sin A sin BProve:
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(60 + 30)
cos (60 + 30) = cos 60 cos 30 – sin 60 sin 30
(1
2)(
3
2) (
3
2)(
1
2) = 0
Aim: Cosine of Two Angles Course: Alg. 2 & Trig.
Model Problem
Find the exact value of cos45ºcos15º – sin45ºsin15º