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Reply 2
I would do cos(3x + 2x)/sin(3x + 2x)
then expand the 3x's using cos(2x + x)/sin(2x + x)
and then expand the 2x's using cos(x + x)/sin(x + x) or using double angle formulae.
there may be a quicker way, but i'm only up to c3
then expand the 3x's using cos(2x + x)/sin(2x + x)
and then expand the 2x's using cos(x + x)/sin(x + x) or using double angle formulae.
there may be a quicker way, but i'm only up to c3
Using Demovire's theorem:
Let c = cosx, s = sinx, then
(c+is)^5
= c^5 + 5c^4(is) + 10c^3(is)^2 + 10c^2(is)^3 + 5c(is)^4 + (is)^5
= c^5 + 5ic^4s - 10c^3s^2 - 10ic^2s^3 + 5cs^4 + is^5
= c^5 - 10c^3s^2 + 5cs^4 + i(5c^4s - 10c^2s^3 + s^5)
collecting the terms free of i gives cos(5x) and collecting terms with i, gives sin(5x) ..
cos(5x) = cos5(x) - 10cos3(x)sin2(x) + 5cos(x)sin4x
sin(5x) = 5cos4(x)sin(x) - 10cos2(x)sin3(x) + sin5(x)
Let c = cosx, s = sinx, then
(c+is)^5
= c^5 + 5c^4(is) + 10c^3(is)^2 + 10c^2(is)^3 + 5c(is)^4 + (is)^5
= c^5 + 5ic^4s - 10c^3s^2 - 10ic^2s^3 + 5cs^4 + is^5
= c^5 - 10c^3s^2 + 5cs^4 + i(5c^4s - 10c^2s^3 + s^5)
collecting the terms free of i gives cos(5x) and collecting terms with i, gives sin(5x) ..
cos(5x) = cos5(x) - 10cos3(x)sin2(x) + 5cos(x)sin4x
sin(5x) = 5cos4(x)sin(x) - 10cos2(x)sin3(x) + sin5(x)
I have a similiar problem i was about to study tomorrow. I remember that in some part of my higher course i had to put an equationthat looked like this
y = BcosCx
into the form y = Dcosx
i'm pretty sure one of the examples was y=2cos2x and the answer turned out to be y = 4cosx or summat...anyone?
y = BcosCx
into the form y = Dcosx
i'm pretty sure one of the examples was y=2cos2x and the answer turned out to be y = 4cosx or summat...anyone?
Reply 5
Mush
I have a similiar problem i was about to study tomorrow. I remember that in some part of my higher course i had to put an equationthat looked like this
y = BcosCx
into the form y = Dcosx
i'm pretty sure one of the examples was y=2cos2x and the answer turned out to be y = 4cosx or summat...anyone?
y = BcosCx
into the form y = Dcosx
i'm pretty sure one of the examples was y=2cos2x and the answer turned out to be y = 4cosx or summat...anyone?
Well you can use double angle formulae to write y = 2cos2x into 3 different ways.
y = 2(cos^2 x - sin^2 x)
y = 2(2 cos^2 x - 1)
y = 2(1 - 2 sin^2 x)
Mush
I have a similiar problem i was about to study tomorrow. I remember that in some part of my higher course i had to put an equationthat looked like this
y = BcosCx
into the form y = Dcosx
i'm pretty sure one of the examples was y=2cos2x and the answer turned out to be y = 4cosx or summat...anyone?
y = BcosCx
into the form y = Dcosx
i'm pretty sure one of the examples was y=2cos2x and the answer turned out to be y = 4cosx or summat...anyone?
If y = 2cos2x, use the identity: cos2x = cos^2(x) - sin^2(x)
y = 2(cos^2(x) - sin^2(x))
= 2(cos^2(x) - (1-cos^2(x)) [*Using]
= 4cos^2(x) - 2
Reply 7
i did the cos(5x) it was a long proses and had taken 3 pages, i came here to find the answer. after doing the question i felt that i should come and display my answer
16Cos(x)^5 - 20Cos(x)^3 + 5Cos(x) = Cos(5x)
16Cos(x)^5 - 20Cos(x)^3 + 5Cos(x) = Cos(5x)
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