Answer to Question #113089 in Trigonometry for Caylin

Question

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) .

Answer 1

$z=\cos A+i\sin A$

the Moivre formula

$z^n=\cos nA+i\sin nA\\ z^{-n}=\frac{1}{z^n}=\frac{1}{\cos nA+i\sin nA}=\\ =\frac{\cos nA-i\sin nA}{(\cos nA+i\sin nA)(\cos nA-i\sin nA)}=\cos nA-i\sin nA$

1)

$z^n+z^{-n}=\\ =\cos nA+i\sin nA+\cos nA-i\sin nA=\\=2\cos nA\\ z^n-z^{-n}=\\ =\cos nA+i\sin nA-\cos nA+i\sin nA=\\=2i\sin nA$

2)

$z+z^{-1}=2\cos A\\ 2\cos A=(z+\frac{1}{z})\\ (2\cos A)^n=(z+\frac{1}{z})^n\\ 2^n\cos^n A=(z+\frac{1}{z})^n\\ \\ z-z^{-1}=2i\sin A\\ 2i\sin A=(z-\frac{1}{z})\\ (2i\sin A)^n=(z-\frac{1}{z})^n\\ (2i)^n\sin^n A=(z-\frac{1}{z})^n$

3)

$\sin^nA=\frac{(z-z^{-1})^n}{(2i)^n}\\ \sin^7A=\frac{(z-z^{-1})^7}{(2i)^7}=\frac{1}{2^7i^7}(z^7-7z^6z^{-1}+\\ +21z^5z^{-2}-35z^4z^{-3}+35z^3z^{-4}-21z^2z^{-5}+\\ +7zz^{-6}-z^{-7})=\\ =\frac{1}{128(-i)}(z^7-z^{-7}-7(z^5-z^{-5})+\\ +21(z^3-z^{-3})-35(z-z^{-1}))=\\ =\frac{i}{128(-i)}(2i\sin7A-7\cdot2i\sin5A+\\ +21\cdot2i\sin3A-35\cdot2i\sin A)=-\frac{2i}{128i}(\sin7A-\\ -7\sin5A+21\sin3A-35\sin A)=\\ =-\frac{1}{64}(\sin7A-7\sin5A+21\sin3A-35\sin A)$

$i^7=i^4\cdot i^3=1\cdot(-i)=-i\\ i^4=(i^2)^2=(-1)^2=1\\ i^3=i^2\cdot i=-1\cdot i=-i$

4)

$\cos^nA=\frac{(z+z^{-1})^n}{2^n}\\ \cos^3A=\frac{(z+z^{-1})^3}{2^3}=\frac{1}{8}(z^3+3z^2z^{-1}+\\ +3zz^{-2}+z^{-3})=\\ =\frac{1}{8}(z^3+z^{-3}+3(z+z^{-1})=\\ =\frac{1}{8}(2\cos3A+3\cdot2\cos A)=\\ =\frac{1}{4}(\cos3A+3\cos A)\\ \sin^nA=\frac{(z-z^{-1})^n}{(2i)^n}\\ \sin^4A=\frac{(z-z^{-1})^4}{(2i)^4}=\frac{1}{16}(z^4-4z^3z^{-1}+\\ +6z^2z^{-2}-4zz^{-3}+z^{-4})=\\ =\frac{1}{16}(z^4+z^{-4}-4(z^2+z^{-2})+6)=\\ =\frac{1}{16}(2\cos4A-4\cdot2\cos 2A+6)=\\ =\frac{1}{8}(\cos4A-4\cos2A+3)$

5)

$4x=\cos 3A+3\cos A\\ \cos^3A=\frac{1}{4}(\cos3A+3\cos A)\\ \cos3A=4\cos^3A-3\cos A\\ 4x=4\cos^3A-3\cos A+3\cos A\\ x=\cos^3A$

$4y=3\sin A-\sin3A\\ \sin^3A=\frac{(z-z^{-1})^3}{(2i)^3}=-\frac{1}{8i}(z^3-3z^2z^{-1}+\\ +3zz^{-2}-z^{-3})=\\ =-\frac{1}{8i}(z^3-z^{-3}-3(z-z^{-1})=\\ =-\frac{1}{8i}(2i\sin3A-3\cdot(2i)\sin A)=\\ =-\frac{1}{4}(\sin3A-3\sin A)\\ \sin 3A=3\sin A-4\sin^3A\\ 4y=3\sin A-3\sin A+4\sin^3A\\ y=\sin^3A$