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The sine of the Sum of 2 Angles:. sin (A + B) = sin A cos B + cos A sin B. Aim: How do we find the trig values of double angles?. Do Now:. Use. to prove sin 2A = 2 sin A cos A. Sine of Double Angles. sin 2A = 2 sin A cos A. sin (A + B) = sin A cos B + cos A sin B. sin 2A = sin(A + A).
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The sine of the Sum of 2 Angles: sin (A + B) = sin A cos B + cos A sin B Aim: How do we find the trig values of double angles? Do Now: Use to prove sin 2A = 2 sin A cos A
Sine of Double Angles sin 2A = 2 sin A cos A sin (A + B) = sin A cos B + cos A sin B sin 2A = sin(A + A) = sin A cos A + cos A sin A sin 2A = 2sin A cos A ex. show that sin 90 = 1 by using sin 2(45). sin 2(45). sin 2(45) = 2 sin 45 cos 45
The Cosine of the Sum of 2 Angles: cos (A + B) = cos A cos B – sin A sin B Prove these two Cosine of Double Angles Prove cos 2A = cos2A – sin2A cos 2A = cos A cos A – sin A sin A cos 2A = cos2 A – sin2 A also cos 2A = 2cos2A – 1 cos 2A = 1 – 2sin2 A and
Prove ex. Show tan 120 by using tan 2(60) Tangent of Double Angles
The Trigonometry of Double Angles sin 2A = 2sin A cos A cos 2A = cos2 A – sin2 A cos 2A = 2cos2A – 1 cos 2A = 1 – 2sin2 A
Model Problems Cos A = -5/13, and A is the measure of an angle in QII. a. Find sin 2A b. Find cos 2A c. Find tan 2A d. Determine the quadrant in which the angle whose measure is 2A lies. sin2 A = 1 – cos2 A a) Find sin A sin2 A = 1 – (-5/13)2 QII - sin A is positive sin2 A = 1 – 25/169 sin2 A = 144/169 sin A = 12/13 sin 2A = 2sin A cos A sin 2A = 2(12/13)(-5/13) = -120/169
Model Problems Cos A = -5/13, and A is the measure of an angle in QII. b. Find cos 2A b) cos 2A = cos2 A – sin2 A cos 2A = (-5/13)2 – (12/13)2 cos 2A = 25/169 – 144/169 cos 2A = - 119/169
Model Problems Cos A = -5/13, and A is the measure of an angle in QII. c. Find tan 2A c) Find tan A
both negative Model Problems Cos A = -5/13, and A is the measure of an angle in QII. d. Determine the quadrant in which the angle whose measure is 2A lies. cos 2A = - 119/169 sin 2A = -120/169 only occurs in QIII
Regents Question The expression is equivalent to (1) csc (3) cot (2) sec (4) sin sin 2A = 2sin A cos A
Regents Question sin 2A = 2sin A cos A sin 2A =
Regents Question cos 2A = cos2 A – sin2 A