Question

prove the identity cos(3x)=cos^3(x)-3sin^2(x)cos(x), step by step

Answer 1

use the formula : cos(A+B) = cosAcosB - sinAsinB
so, cos(3x) = cos(x+2x) = ....

Answer 2

cos(x)*cos(2x)-sin(x)*sin(2x)

Answer 3

yes, so far is good.
now, use the identity that cos2x = 2cos^2 (x) - 1 and
sin(2x) = 2sin(x)cos(x)
apply of them to did u get above

Answer 4

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

Answer 5

then simplify it ...
cos(x)cos^2 (x) = ....
sin(x)*2sin(x)cos(x) = ...

Answer 6

cos^3(x)-1-3sin^2(x)cos(x)

Answer 7

what about the 1?

Answer 8

If you know Euler's formula:\[e^{i\theta}=\cos(\theta)+i\sin(\theta)\]there's another way to do this.

Answer 9

i need to make a proof, and dont know that formula.

Answer 10

You don't get the 1. You forgot to multiply the one with cosx

Answer 11

\[\cos(x)[\cos^{2}(x)-1]-\sin(x)[2\sin(x)\cos(x)]\]

Answer 12

\[2cos^2(x)\] Made a mistake in the brackets of the first term.

Answer 13

im not sure what im missing, but what cancels the 1?

Answer 14

@comput313 I just told you. Why don't you understand what I wrote.

Answer 15

expand this part first. Tell me what you get.
\[cosx(2cos^{2}x-1)\]

Answer 16

After you expanded that. Expand this:
\[sinx(2sinxcosx)=?\]

Answer 17

cal me an idiot, but i still dont see what removes the 1.
cos(x)*cos^2(x)-1=cos^3(x)-1

Answer 18

You need to learn how to expand properly.

Answer 19

expand this. x(x+1)=

Answer 20

what you did above was this.
x(x+1)=x^2+1. That's not how you expand.

Answer 21

This is how you expand. x(x+1)=x^2+x

Answer 22

i see now, cos(x)*cos^2(x)-1= 2cos^3(x)- 1cos(x) =cos^3(x)

Answer 23

Now. We're not finished yet.

Answer 24

factorise the cosx out.

Answer 25

And that's incorrect.

Answer 26

2cos^3(x)- 1cos(x) does not equal cos^3(x)

Answer 27

@comput313

Answer 28

this is why i have a bad grade in math. please continue explaining.

Answer 29

Factorise the cosx out.
\[2\cos^{3}x-cosx-2\sin^{2}xcosx=?\]

Answer 30

\[=cosx[2\cos^{2}x-1-2\sin^{2}x]\]
You get that right?

Answer 31

Now You must know this trig equation.
\[\sin^{2}x+\cos^{2}x=1\]

Answer 32

no, let me work through it again

Answer 33

yes i know that

Answer 34

Fine you use the double angle results.
\[\cos3x=\cos(x+2x)\]
\[\cos(x+2x)=\cos(x)\cos2x-\sin(x)\sin2x\]

Answer 35

no, i got that part

Answer 36

i factored wrong i got it now

Answer 37

\[\cos(x)\cos2x-\sin(x)\sin2x=\cos(x)(\cos^{2}x-sin^{2}x)-\sin(x)(2\sin(x)\cos(x))\]

Answer 38

Okay.

Answer 39

We go back to using that trig equation.
\[\sin^{2}x+\cos^{2}x=1\]

Answer 40

So everytime we see 1, we substitute \[\sin^{2}x+\cos^{2}x\]

Answer 41

ah, i see now

Answer 42

\[\cos(x)(2\cos^{2}x-1-2\sin^{2}x)\]
\[=\cos(x)(2\cos^{2}x-(\sin^{2}x+\cos^{2}x)-2\sin^{2}x)\]

Answer 43

Now collect like terms and expand again. Once you do that, it will equal the RHS(Right Hand Side) and thus you proved it.

Answer 44

well, some formulas i gave above doesnt work...
try use the identity cos2x = cos^2 x - sin^2 x, like aztech said
let's try again...
cos(x+2x) = cosxcos2x - sinxsin2x
cos3x = cosx(cos^2 x -sin^2 x) - sinx(2sinxcosx)
cos3x = cos^3 x - sin^2 x cosx - 2sin^2 x cosx
cos3x = cos^3 x - 3sin^2 x cosx
proof

Answer 45

remember
\[-(x+y)=-x-y\]
\[-(x+y)\neq -x+y\]
I see many students try and do that.

Answer 46

ok, thank you both

Answer 47

yw