Math, asked by ashuramani, 4 months ago

derive multiple angle formula​

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Answered by Anonymous
2

Answer:

For n a positive integer, expressions of the form sin(nx), cos(nx), and tan(nx) can be expressed in terms of sinx and cosx only using the Euler formula and binomial theorem.

For sin(nx),

sin(nx) = (e^(inx)-e^(-inx))/(2i)

(1)

= ((e^(ix))^n-(e^(-ix))^n)/(2i)

(2)

= ((cosx+isinx)^n-(cosx-isinx)^n)/(2i)

(3)

= sum_(k=0)^(n)(n; k)(cos^kx(isinx)^(n-k)-cos^kx(-isinx)^(n-k))/(2i)

(4)

= sum_(k=0)^(n)(n; k)cos^kxsin^(n-k)x(i^(n-k)-(-i)^(n-k))/(2i)

(5)

= sum_(k=0)^(n)(n; k)cos^kxsin^(n-k)xsin[1/2(n-k)pi].

(6)

The first few values are given by

sin(2x) = 2cosxsinx

(7)

sin(3x) = 3cos^2xsinx-sin^3x

(8)

sin(4x) = 4cos^3xsinx-4cosxsin^3x

(9)

sin(5x) = 5cos^4xsinx-10cos^2xsin^3x+sin^5x.

(10)

Other related formulas include

sin(nx) = sinxsum_(k=0)^(|_(n-1)/2_|)(-1)^k(n-k-1; k)2^(n-2k-1)cos^(n-2k-1)x

(11)

= sum_(k=0)^(|_(n-1)/2_|)(-1)^k(n; 2k+1)sin^(2k+1)xcos^(n-2k-1)x,

(12)

where |_x_| is the floor function.

A product formula for sin(nx) is given by

sin(nx)=2^(n-1)product_(k=0)^(n-1)sin((pik)/n+x).

(13)

The function sin(nx) can also be expressed as a polynomial in sinx (for n odd) or cosx times a polynomial in sinx as

sin(nx)={(-1)^((n-1)/2)T_n(sinx) for n odd; (-1)^(n/2-1)cosxU_(n-1)(sinx) for n even,

(14)

where T_n is a Chebyshev polynomial of the first kind and U_n is a Chebyshev polynomial of the second kind. The first few cases are

sin(2x) = 2cosxsinx

(15)

sin(3x) = 3sinx-4sin^3x

(16)

sin(4x) = cosx(4sinx-8sin^3x)

(17)

sin(5x) = 5sinx-20sin^3x+16sin^5x.

(18)

Similarly, sin(nx) can be expressed as sinx times a polynomial in cosx as

sin(nx)=sinxU_(n-1)(cosx).

(19)

The first few cases are

sin(2x) = 2cosxsinx

(20)

sin(3x) = sinx(-1+4cos^2x)

(21)

sin(4x) = sinx(-4cosx+8cos^3x)

(22)

sin(5x) = sinx(1-12cos^2x+16cos^4x).

(23)

Bromwich (1991) gave the formula

sin(na)={nx-(n(n^2-1^2)x^3)/(3!)+(n(n^2-1^2)(n^2-3^2)x^5)/(5!)-... for n odd; ncosa[x-((n^2-2^2)x^3)/(3!)+((n^2-2^2)(n^2-4^2)x^5)/(5!)-...] for n even,

(24)

where x=sina.

For cos(nx), the multiple-angle formula can be derived as

cos(nx) = (e^(inx)+e^(-inx))/2

(25)

= ((e^(ix))^n+(e^(-ix))^n)/2

(26)

= ((cosx+isinx)^n+(cosx-isinx)^n)/2

(27)

= sum_(k=0)^(n)(n; k)(cos^kx(isinx)^(n-k)+cos^kx(-isinx)^(n-k))/2

(28)

= sum_(k=0)^(n)(n; k)cos^kxsin^(n-k)x(i^(n-k)+(-i)^(n-k))/2

(29)

= sum_(k=0)^(n)(n; k)cos^kxsin^(n-k)xcos[1/2(n-k)pi].

(30)

The first few values are

cos(2x) = cos^2x-sin^2x

(31)

cos(3x) = cos^3x-3cosxsin^2x

(32)

cos(4x) = cos^4x-6cos^2xsin^2x+sin^4x

(33)

cos(5x) = cos^5x-10cos^3xsin^2x+5cosxsin^4x.

(34)

Other related formulas include

cos(nx) = nsum_(k=0)^(|_n/2_|)((-1)^k(n-k-1)!2^(n-2k-1)cos^(n-2k)x)/(k!(n-2k!))

(35)

= 2^(n-1)cos^nx+nsum_(k=1)^(|_n/2_|)((-1)^k)/k(n-k-1; k-1)2^(n-2k-1)cos^(n-2k)x

(36)

= sum_(k=0)^(|_n/2_|)(-1)^k(n; 2k)sin^(2k)xcos^(n-2k)x.

(37)

The function cos(nx) can also be expressed as a polynomial in sinx (for n even) or cosx times a polynomial in sinx as

cos(nx)={(-1)^((n-1)/2)cosxU_(n-1)(sinx) for n odd; (-1)^(n/2)T_n(sinx) for n even.

(38)

The first few cases are

cos(2x) = 1-2sin^2x

(39)

cos(3x) = cosx(1-4sin^2x)

(40)

cos(4x) = 1-8sin^2x+8sin^4x

(41)

cos(5x) = cosx(1-12sin^2x+16sin^4x).

(42)

Similarly, cos(nx) can be expressed as a polynomial in cosx as

cos(nx)=T_n(cosx).

(43)

The first few cases are

cos(2x) = -1+2cos^2x

(44)

cos(3x) = -3cosx+4cos^3x

(45)

cos(4x) = 1-8cos^2x+8cos^4x

(46)

cos(5x) = 5cosx-20cos^3x+16cos^5x.

(47)

Bromwich (1991) gave the formula

cos(na)={cosa[1-((n^2-1^2)x^2)/(2!)+((n^2-1^2)(n^2-3^2)x^4)/(4!)-...] n odd; 1-(n^2x^2)/(2!)+(n^2(n^2-2^2)x^4)/(4!)-... n even,

(48)

where x=sina.

The first few multiple-angle formulas for tan(nx) are

tan(2x) = (2tanx)/(1-tan^2x)

(49)

tan(3x) = (3tanx-tan^3x)/(1-3tan^2x)

(50)

tan(4x) = (4tanx-4tan^3x)/(1-6tan^2x+tan^4x)

(51)

are given by Beyer (1987, p. 139) for up to n=6.

Multiple-angle formulas can also be written using the recurrence relations

sin(nx) = 2sin[(n-1)x]cosx-sin[(n-2)x]

(52)

cos(nx) = 2cos[(n-1)x]cosx-cos[(n-2)x]

(53)

tan(nx) = (tan[(n-1)x]+tanx)/(1-tan[(n-1)x]tanx).

Step-by-step explanation:

hope it helps

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