Question

Sum and difference identities : prove cos3x=cos^3x-3sin^2xcosx

Answer 1

I understand how to prove most of these but the exponents are throwing me off , could someone walk me through step by step please? Thank you!

Answer 2

try usong cos (2 + x) = cos2x cos x - sin2x sinx

Answer 3

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

Answer 4

= 2 cos^3 x - cosx - 2 sin^2x cosx

Answer 5

hmm - doesn''t look like we are going anywhere with this

Answer 6

@welshfella it looks good to me
now you need double angle identities

Answer 7

oh you did that I didn't read all the posts

Answer 8

\[(\cos^2(x)-\sin^2(x))\cos(x)-2\sin(x)\cos(x)\sin(x)\]

Answer 9

@bastiennecroix can you try to take it from here

Answer 10

I originally shied away from this problem because of the ambiguity in your problem statement: cos3x=cos^3x-3sin^2xcosx.

Does your "cos 3x" mean just that, or does it mean (cos x)^3?

It may be worth your while to learn how to use Equation Editor, or at least the Draw utility, below:

\[\cos 3x=\cos^3x-3\sin^2xcos x\]

Answer 11

Please verify that this re-typed statement is correct (or not).

Answer 12

hint:
if we apply the formula of Euler, we can write:
\[\Large {e^{ix}} = \cos x + i\sin x\]
from which we get:
\[\Large {e^{i3x}} = \cos \left( {3x} \right) + i\sin \left( {3x} \right)\]
furthermore, we have:
\[\Large {e^{i3x}} = {\left( {{e^{ix}}} \right)^3} = {\left( {\cos x + i\sin x} \right)^3}\]
so we get thie equality:
\[\Large \cos \left( {3x} \right) + i\sin \left( {3x} \right) = {\left( {\cos x + i\sin x} \right)^3}\]
please develop the third power at the right side, and equate the real parts of both sides

Answer 13

ah - yes that might be the way to go

= cos^3 x - sin^2 x cosx - 2 sin^2 x cos x

Answer 14

thats it
I picked the wrong identity for cos 2x

Answer 15

You may want to use the sum formula for the cosine:

cos (x+y) = cos x cos y - sin x sin y

Answer 16

...so that cos 3x = cos (x + 2x)= ??? Please expand this. Decide for yourself whether your result will help you answer your original question.