Double-angle & powe-eduction identities Pat 5, Tigonomety Lectue 5a, Double Angle and Powe Reduction Fomulas In the pevious pesentation we developed fomulas fo cos( β) and sin( β) These fomulas lead natually to anothe set of identities involving double angles powe eduction and half-angles D Ken W Smith Sam Houston State Univesity 03 03 / 7 Double-angle & powe-eduction identities 03 / 7 Double-angle & powe-eduction identities Recall the sum-of-angle identities fom an ealie pesentation: The equation cos( β) cos cos β sin sin β cos(θ) cos θ sin θ is paticulaly applicable sin( β) sin cos β cos sin β If we eplace and β with the same angle, θ, these identities descibe the sine and cosine of θ, expessing tig functions of a doubled angle in tems of the oiginal cos(θ) cos θ sin θ () sin(θ) sin θ cos θ It can be used to educe the powe on cosine and sine Fo example, suppose we add this equation to the Pythagoean Identity cos θ sin θ () 03 3 / 7 03 4 / 7
Double-angle & powe-eduction identities Double-angle & powe-eduction identities We get cos θ cos θ sin θ cos θ sin θ cos θ cos θ Afte dividing by, we obtain an equation fo cos θ cos θ ( cos θ) (3) In a simila manne, we could subtact the double angle fomula fom the Pythagoean identity: cos θ sin θ getting sin θ cos θ (cos θ cos θ sin θ) Afte dividing both sides by we have a fomula fo sin θ: sin θ ( cos θ) (4) 03 5 / 7 Double-angle & powe-eduction identities 03 6 / 7 Sum and Diffeence Fomulas cos θ ( cos θ) sin θ ( cos θ) These fomulas ae sometimes called powe eduction fomulas because they often allow us educe the powe on one of the tig functions, if the powe is an even intege Fo example, we can educe cos 4 x (cos x) by substituting fo cos x and then expanding the expession In the next pesentation we look at half-angle fomulas (End) cos 4 x ( cos x ) 4 ( cos xcos x) cos 4x ( cos x( )) 4 We may simplify the last expession by using the common denominato 8 and so wite: cos 4 x (3 4 cos x cos 4x)) 8 03 7 / 7 03 8 / 7
Half-angle fomulas The powe eduction fomulas can also be intepeted as half-angle fomulas Conside again the powe eduction fomulas cos θ ( cos θ) and sin θ ( cos θ) Replace θ by (and eplace θ by ) Now ou equations ae Pat 5, Tigonomety Lectue 5b, Half-angle Fomulas cos ( ) ( cos ) and sin ( ) ( cos ) Now take squae oots of both sides: cos( ) ± ( cos ) sin( ) ± ( cos ) D Ken W Smith Sam Houston State Univesity 03 03 9 / 7 Half-angle fomulas (6) 03 0 / 7 Some woked poblems cos( ) ± ( cos ) ( cos ) sin( ) ± π Use half-angle fomulas to find the exact value of cos Solution We view π/ as half of the angle π/6 So using the half-angle fomula fo cosine, with π/6, we have p cos(π/6) 3/ 3 3 π cos( ) 4 If we want an equation fo tan( ) we divide sin by cos to get cos tan( ) ± cos (5) (7) 03 / 7 03 / 7
Some woked poblems Some woked poblems Use half-angle fomulas to find the exact value of sin π Solution Again we view π/ as half of the angle π/6 and hee we use half-angle fomula fo sine cos(π/6) 3/ 3 3 sin( π ) 4 Use half-angle fomulas to find the exact value of cos π 4 Solution The angle π/4 is half of the angle π/ Fotunately we just computed the cosine and sine of π/ so we can use the half-angle fomula again: cos( π 3 cos(π/) 4 ) 3 4 3 03 3 / 7 Some woked poblems 03 4 / 7 An inteesting question Use half-angle fomulas to find the exact value of sin π 4 Solution sin( π 4 ) cos(π/) 3 3 Is thee a patten to the fomula fo sine and cosine if we stat at π/6 and keep cutting the angle in half? sin( π ) 3 sin( π 4 ) 3 Since the sine function helps give us a chod on the cicle, we could then develop a fomula fo the cicumfeence of a cicle and then develop, fom that, a fomula fo π Achimedes did this by (essentially) computing sin( π 96 )! 03 5 / 7 03 6 / 7
Sum and Diffeence Fomulas In the next pesentation we look at the Law of Sines (End) 03 7 / 7