Sin(10x) = sinx(512(cos^9x) - 1024(cos^7x) + 672(cos^5x) - 160(cos^3x) + 10cosx)?

Question

Prove Identity.

Yes i know isnt this the most rediculous problem?

Yes it really is Sin(10x)

Answer 1

You can prove your identity in this way.

sin2x = 2 sinx cosx

cos2x = cos^2x - sin^2x
= cos^2x - (1 - cos^2x)
= cos^2x - 1 + cos^2x
= 2 cos^2x - 1

sin3x = sin(2x + x)
= sin2x cox + cos2x sinx
(Using the formula sin (a+b) = sina cosb + cosa sinb)
= (2 sinx cosx) cosx + (2 cos^2x - 1) sinx
= 2 sinx cos^2x + 2 cos^2x sinx - sinx
= 4 cos^2x sinx - sinx

cos3x = cos(2x + x)
= cos2x cosx - sin2x sinx
(Using the formula cos (a+b) = cosa cosb - sina sinb)
= (2 cos^2x - 1) cosx - (2 sinx cosx) sinx
= 2 cos^3x - cosx - 2 sin^2x cosx
= 2 cos^3x - cosx - 2 (1 - cos^2x) cosx
= 2 cos^3x - cosx - 2 cosx + 2 cos^3x
= 4 cos^3x - 3 cosx

sin5x = sin (3x + 2x) = sin3x cos2x + cos3x sin2x
= (4 cos^2x sinx - sinx) (2 cos^2x - 1) + (4 cos^3x - 3 cox) (2 sinx cosx)
= 8 cos^4x sinx - 4 cos^2x sinx - 2 cos^2x sinx + sinx + 8 cos^4x sinx - 6 cos^2x sinx
= 16 sinx cos^4x - 12 sinx cos^2x + sinx

cos5x = cos (3x + 2x) = cos3x cos2x - sin3x sin2x
= (4 cos^3x - 3 cosx) ( 2 cos^2x -1) - (4 cos^2x sinx - sinx) (2 sinx cosx)
= 8 cos^5x - 4 cos^3x - 6 cos^3x + 3 cosx - 8 cos^3x sin^2x + 2 sin^2x cos x
= 8 cos^5x - 10 cos^3x + 3 cosx - 8 cos^3x (1 - cos^2x) + 2 (1 - cos^2x) cosx
= 8 cos^5x - 10 cos^3x + 3 cosx - 8 cos^3x + 8 cos^5x + 2 cosx - 2 cos^3x
= 16 cos^5x - 20 cos^3x + 5 cosx

sin10x = 2 sin5x cos5x
= 2 (16 sinx cos^4x - 12 sinx cos^2x + sinx) (16 cos^5x - 20 cos^3x + 5 cosx)
= 2 (256 sinx cos^9x - 320 sinx cos^7x + 80 sinx cos^5x - 192 sinx cos^7x + 240 sinx cos^5x - 60 sinx cos^3x + 16 sinx cos^5x - 20 sinx cos^3x + 5 sinx cosx)
= 2 (256 sinx cos^9x - 512 sinx cos^7x + 336 sinx cos^5x - 80 sinx cos^3x + 5 sinx cosx)
= sinx(512 cos^9x - 1024 cos^7x + 672 cos^5x - 160 cos^3x + 10 cosx)

I know this looks really lengthy but i have tried to make it as simple as possible.

Hope this helps.

Answer 2

Proof your equation. Do you really mean Sin(10x)? I suspect that whatever the identity winds up as, sin^2(x)+cos^2(x) = 1 will have something to do with it, as will 2sinxcosx = sin 2x. You can factor out one of these on the right to get
Sin 10x ?= sin 2x (256 cos^8 x - 512 cos^6 x + 336 cos^4 x - 80 cos^2 x + 5)

Answer 3

left hand side = sin 10 x and RHS except for sin x is power of cos x and sum/diff of them

that induces us to use the following form

e^i 10 x = cos 10 x + i sin 10x
e^-i 10 x = cos 10 x - i sin 10x
subtract and get

2 i sin 10 x = (cos x + i sin x ) ^ 10 - ( cos x - i sin x )^10

you can now expand the RHS and put sin ^2 x = 1- cos^2 x and get the result

Answer 4

Hint:

e^(i10x) = [cos(10x) + isin(10x)] = [cos(x) + isin(x)]^10

Then compare the coefficients of imaginary parts.

Answer 5

yourr teacher is a masochist!