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MCCC MAT 146 - Sums and Differences of Angles

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Sums and Differences of Angles MAT146 FALL 2011cos (α+β) = cos α cos β − sin α sin β • First:Then: • The angle β with terminal points at Q (cos α, sin α) and R (cos (α + β), sin (α + β)) • The angle -β with terminal point at S (cos (-β), sin (-β)) • The lines PR and QS, which are equivalent in length.Here’s the picture: Note:The Distance Formula: • Using Pythagoras' Theorem we have the distance between (x1, y1) and (x2, y2) given by:PR2 = (cos (α + β) − 1)2 + sin2(α + β) = cos2(α + β) − 2cos (α + β) + 1 + sin2(α + β) = 2 − 2cos (α + β) [since sin2(α + β) + cos2(α + β) = 1]QS² = (cos α − cos (-β))² + (sin α − sin (-β))² = cos²α − 2cos α cos(-β) + cos²(-β) + sin ²α − 2sin α sin(-β) + sin²(-β) = 2 − 2cos α cos(-β) − 2sin α sin(-β) [since sin²α + cos²α = 1 and sin²(-β) + cos²(-β) = 1] = 2 − 2cos α cos β + 2sin α sin β [since cos(-β) = cos β (cosine is an even function) and sin(-β) = -sinβ (sine is an odd functionSince PR = QS, we can equate the 2 distances we just found: • 2 − 2cos (α + β) = 2 − 2cos α cos β + 2sin α sin βSubtracting 2 from both sides and dividing throughout by -2, we obtain: cos (α + β) = cos α cos β − sin α sin β If we replace β with (-β), this identity becomes: cos (α − β) = cos α cos β + sin α sin β [since cos(-β) = cos βand sin(-β) = -sinβ]Sin(ά+β) • Recall that sin (θ) = cos (π/2− θ) • If θ = α + β, then we have: sin (α + β) = cos [π/2 − (α + β)] = cos [π/2 − α − β)]• Regroup: cos [π/2 − α − β)] = cos [(π/2 − α) − β] Using the cosine of the difference of 2 angles identity that we just found : cos [(π/2 − α) − β] = cos (π/2 − α) cos (β) + sin (π/2 − α) sin (β) = sin α cos β + cos α sin β Since cos (π/2 − α) = sin α; and sin (π/2 − α) = cos α]More: • Replacing β with (-β), this identity becomes (because of even and odd functions): sin (α − β) = sin α cos β − cos α sin


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