Lecture 20:Trigonometric FunctionsDan SloughterFurman UniversityMathematics 39April 7, 200420.1 Defining sine and cosineRecall that if x ∈ R, thenex=∞Xk=0xkk!= 1 + x +x22+x33!+x44!+ · · · .Note what happens if we (somewhat blindly) let x = iθ:eiθ= 1 + iθ −θ22− iθ33!+θ44!+ iθ55!−θ66!− iθ77!+ · · ·=1 −θ22+θ44!−θ66!+ · · ·+ iθ −θ33!+θ55!−θ77!+ · · ·= cos(θ) + i sin(θ ).This is the motivation for our earlier definition of eiθ. It now follows that forany x ∈ R, we haveeix= cos(x) + i sin(x) and e−ix= cos(x) − i sin(x),from which we obtain (by addition)2 cos(x) = eix+ e−ixand (by subtraction)2i sin(x) = eix− eix.1Hence we havecos(x) =eix+ e−ix2andsin(x) =eix− e−ix2,which motivate the following definitions.Definition 20.1. For any c omplex number z, we define the sine function bysin(z) =eiz− e−iz2iand the cosine function bycos(z) =eiz+ e−iz2.The f ollowing proposition is immediate from the properties of analyticfunctions and the fact that ezis an entire function.Proposition 20.1. Both sin(z) and cos(z) are entire functions.Proposition 20.2. For all z ∈ C,ddzsin(z) = cos(z) andddzcos(z) = − sin(z).Proof. We haveddzsin(z) =ieiz+ ie−iz2i= cos(z)andddzcos(z) =ieiz− ie−iz2= −eiz− e−iz2i= − sin(z).20.2 Properties of sine and cosineProposition 20.3. For any z ∈ C,sin(−z) = − sin(z) and cos(−z) = cos(z).2Proof. We havesin(−z) =e−iz− eiz2i= − sin(z)andcos(−z) =e−iz+ eiz2= cos(z).Proposition 20.4. For any z1, z2∈ C,2 sin(z1) cos(z2) = sin(z1+ z2) + sin(z1− z2).Proof. We have2 sin(z1) cos(z2) = 2eiz1− e−iz12ieiz2+ e−iz22=ei(z1+z2)+ ei(z1−z2)− e−i(z1−z2)− e−i(z1+z2)2i=ei(z1+z2)− e−i(z1+z2)2i+ei(z1−z2)− e−i(z1−z2)2i= sin(z1+ z2) + sin(z1− z2).Proposition 20.5. For any z1, z1∈ C,sin(z1+ z2) = sin(z1) cos(z2) + cos(z1) sin(z2)andcos(z1+ z2) = cos(z1) cos(z2) − sin(z1) sin(z2).Proof. From the previous result, we have2 sin(z1) cos(z2) = sin(z1+ z2) + sin(z1− z2)and2 sin(z2) cos(z1) = sin(z1+ z2) − sin(z1− z2),from which we obtain the first identify by addition. It now follows that iff(z) = sin(z + z2),3thenf(z) = sin(z) cos(z2) + cos(z) sin(z2)as well. Hencecos(z1+ z2) = f0(z1) = cos(z1) cos(z2) − sin(z1) sin(z2).The following identities follow immediately from the previous proposi-tions.Proposition 20.6. For any z ∈ C,sin2(z) + cos2(z) = 1,sin(2z) = 2 sin(z) cos(z),cos(2z) = 2 cos2(z) − sin2(z),sinz +π2= cos(z),sinz −π2= − cos(z),cosz +π2= − sin(z),cosz −π2= sin(z),sin(z + π) = − sin(z),cos(z + π) = − cos(z),sin(z + 2π) = sin(z),andcos(z + 2π) = cos(z).Proposition 20.7. For any z = x + iy ∈ C,sin(z) = sin(x) cosh(y) + i cos(x) sinh(y)andcos(z) = cos(x) cosh(y) − i sin(x) sinh(y).4Proof. We first note thatcos(iy) =e−y+ ey2= cosh(y)andsin(iy) =e−y− ey2i= iey− e−y2= i sinh(y).Hencesin(x + iy) = sin(x) cos(iy) + sin(iy) cos(x) = sin(x) cosh(y) + i cos(x) sinh(y)andcos(x+iy) = cos(x) cos(iy)−sin(x) sin(iy) = cos(x) cosh(y)−i sin(x) sinh(y).It now follows (see the homework)that| sin(z)|2= sin2(x) + sinh2(y)and| cos(z)|2= cos2(x) + sinh2(y).Since sinh(y) = 0 if and only if y = 0, we see that sin(z) = 0 if and only ify = 0 and x = nπ for some n = 0, ±1, ±2, . . ., and cos(z) = 0 if and only ify = 0 and x =π2+ nπ for some n = 0, ±1, ±2, . . .. That is, sin(z) = 0 if andonly ifz = nπ, n = 0, ±1, ±2, . . . ,and cos(z) = 0 if and only ifz =π2+ nπ, n = 0, ±1, ±2, . . . .20.3 The other trigonometric functionsThe rest of the trigonometric functions are defined as usual:tan(z) =sin(z)cos(z),5cot(z) =cos(z)sin(z),sec(z) =1cos(z),andcsc(z) =1sin(z).Using our results on derivatives, it is straightforward to show thatddztan(z) = sec2(z),ddzcot(z) = − csc2(z),ddzsec(z) = sec(z) tan(z),andddzcsc(z) = − csc(z) cot(z).In particular, these functions are analytic at all points at which they aredefined. As with their real counterparts, they are all periodic, tan(z) andcot(z) having period π and sec(z) and csc(z) having period