Question #113089
Let z = cosA + sinA

then z^n = cos(nA) + i sin(nA) for all n E N (by de Moivre) and z^-n = cos(nA) - i sin(nA).

1) show that 2cos(nA) = z^n + z^-n and 2i sin(nA) = z^n - z^-n.

2) show that 2^n cos^n A = ( z + 1/z)^n and (2i)^n sin ^n A= (z - 1/z)^n.

3) Use (2) to express sin^7 A in terms of multiple angles

4) Express cos^3 A sin ^4 A in terms of multiple angles

5) Eliminate A from the equations 4x = cos(3A) + 3cosA ; 4y = 3 sin A - sin (3A) .
1
Expert's answer
2020-04-30T19:07:11-0400

z=cosA+isinAz=\cos A+i\sin A

the Moivre formula

zn=cosnA+isinnAzn=1zn=1cosnA+isinnA==cosnAisinnA(cosnA+isinnA)(cosnAisinnA)=cosnAisinnAz^n=\cos nA+i\sin nA\\ z^{-n}=\frac{1}{z^n}=\frac{1}{\cos nA+i\sin nA}=\\ =\frac{\cos nA-i\sin nA}{(\cos nA+i\sin nA)(\cos nA-i\sin nA)}=\cos nA-i\sin nA

1)

zn+zn==cosnA+isinnA+cosnAisinnA==2cosnAznzn==cosnA+isinnAcosnA+isinnA==2isinnAz^n+z^{-n}=\\ =\cos nA+i\sin nA+\cos nA-i\sin nA=\\=2\cos nA\\ z^n-z^{-n}=\\ =\cos nA+i\sin nA-\cos nA+i\sin nA=\\=2i\sin nA

2)

z+z1=2cosA2cosA=(z+1z)(2cosA)n=(z+1z)n2ncosnA=(z+1z)nzz1=2isinA2isinA=(z1z)(2isinA)n=(z1z)n(2i)nsinnA=(z1z)nz+z^{-1}=2\cos A\\ 2\cos A=(z+\frac{1}{z})\\ (2\cos A)^n=(z+\frac{1}{z})^n\\ 2^n\cos^n A=(z+\frac{1}{z})^n\\ \\ z-z^{-1}=2i\sin A\\ 2i\sin A=(z-\frac{1}{z})\\ (2i\sin A)^n=(z-\frac{1}{z})^n\\ (2i)^n\sin^n A=(z-\frac{1}{z})^n

3)

sinnA=(zz1)n(2i)nsin7A=(zz1)7(2i)7=127i7(z77z6z1++21z5z235z4z3+35z3z421z2z5++7zz6z7)==1128(i)(z7z77(z5z5)++21(z3z3)35(zz1))==i128(i)(2isin7A72isin5A++212isin3A352isinA)=2i128i(sin7A7sin5A+21sin3A35sinA)==164(sin7A7sin5A+21sin3A35sinA)\sin^nA=\frac{(z-z^{-1})^n}{(2i)^n}\\ \sin^7A=\frac{(z-z^{-1})^7}{(2i)^7}=\frac{1}{2^7i^7}(z^7-7z^6z^{-1}+\\ +21z^5z^{-2}-35z^4z^{-3}+35z^3z^{-4}-21z^2z^{-5}+\\ +7zz^{-6}-z^{-7})=\\ =\frac{1}{128(-i)}(z^7-z^{-7}-7(z^5-z^{-5})+\\ +21(z^3-z^{-3})-35(z-z^{-1}))=\\ =\frac{i}{128(-i)}(2i\sin7A-7\cdot2i\sin5A+\\ +21\cdot2i\sin3A-35\cdot2i\sin A)=-\frac{2i}{128i}(\sin7A-\\ -7\sin5A+21\sin3A-35\sin A)=\\ =-\frac{1}{64}(\sin7A-7\sin5A+21\sin3A-35\sin A)

i7=i4i3=1(i)=ii4=(i2)2=(1)2=1i3=i2i=1i=ii^7=i^4\cdot i^3=1\cdot(-i)=-i\\ i^4=(i^2)^2=(-1)^2=1\\ i^3=i^2\cdot i=-1\cdot i=-i

4)

cosnA=(z+z1)n2ncos3A=(z+z1)323=18(z3+3z2z1++3zz2+z3)==18(z3+z3+3(z+z1)==18(2cos3A+32cosA)==14(cos3A+3cosA)sinnA=(zz1)n(2i)nsin4A=(zz1)4(2i)4=116(z44z3z1++6z2z24zz3+z4)==116(z4+z44(z2+z2)+6)==116(2cos4A42cos2A+6)==18(cos4A4cos2A+3)\cos^nA=\frac{(z+z^{-1})^n}{2^n}\\ \cos^3A=\frac{(z+z^{-1})^3}{2^3}=\frac{1}{8}(z^3+3z^2z^{-1}+\\ +3zz^{-2}+z^{-3})=\\ =\frac{1}{8}(z^3+z^{-3}+3(z+z^{-1})=\\ =\frac{1}{8}(2\cos3A+3\cdot2\cos A)=\\ =\frac{1}{4}(\cos3A+3\cos A)\\ \sin^nA=\frac{(z-z^{-1})^n}{(2i)^n}\\ \sin^4A=\frac{(z-z^{-1})^4}{(2i)^4}=\frac{1}{16}(z^4-4z^3z^{-1}+\\ +6z^2z^{-2}-4zz^{-3}+z^{-4})=\\ =\frac{1}{16}(z^4+z^{-4}-4(z^2+z^{-2})+6)=\\ =\frac{1}{16}(2\cos4A-4\cdot2\cos 2A+6)=\\ =\frac{1}{8}(\cos4A-4\cos2A+3)


5)

4x=cos3A+3cosAcos3A=14(cos3A+3cosA)cos3A=4cos3A3cosA4x=4cos3A3cosA+3cosAx=cos3A4x=\cos 3A+3\cos A\\ \cos^3A=\frac{1}{4}(\cos3A+3\cos A)\\ \cos3A=4\cos^3A-3\cos A\\ 4x=4\cos^3A-3\cos A+3\cos A\\ x=\cos^3A


4y=3sinAsin3Asin3A=(zz1)3(2i)3=18i(z33z2z1++3zz2z3)==18i(z3z33(zz1)==18i(2isin3A3(2i)sinA)==14(sin3A3sinA)sin3A=3sinA4sin3A4y=3sinA3sinA+4sin3Ay=sin3A4y=3\sin A-\sin3A\\ \sin^3A=\frac{(z-z^{-1})^3}{(2i)^3}=-\frac{1}{8i}(z^3-3z^2z^{-1}+\\ +3zz^{-2}-z^{-3})=\\ =-\frac{1}{8i}(z^3-z^{-3}-3(z-z^{-1})=\\ =-\frac{1}{8i}(2i\sin3A-3\cdot(2i)\sin A)=\\ =-\frac{1}{4}(\sin3A-3\sin A)\\ \sin 3A=3\sin A-4\sin^3A\\ 4y=3\sin A-3\sin A+4\sin^3A\\ y=\sin^3A




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