Question

How to start this proving this equation? cos(3x)=4COS^3x-3COSx

Answer 1

*How to start proving this equation?

Answer 2

use the addition angle formula for cosine twice

Answer 3

\[\cos(3x)=\cos(2x+x)\] for the first step

Answer 4

Do you mean sum and difference identity for cosine?

Answer 5

I don't know the addition angle formula. Is there another way of proving this identity?

Answer 6

use the formula :
cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
here, let A=2x and B=x

Answer 7

what @RadEn said

Answer 8

then when you get
\[\cos(2x)\sin(x)-\sin(2x)\sin(x)\] you can use the double angle formula for \(\sin(2x)\) and \(\cos(2x)\)

Answer 9

How are you getting 2x?

Answer 10

3x=2x+x, right ?

Answer 11

yes

Answer 12

oh. Are ya working from the left side of the problem?

Answer 13

yes, but we have to simplify for right side with identities trigono

Answer 14

we have :
cos(3x) = cos(2x+x)
cos(3x) = cos(2x)cos(x) - sin(2x)sin(x)

Answer 15

now, use the identities :
cos(2x) = 2cos^2 (x) and
sin(2x) = 2sin(x)cos(x)

Answer 16

so, it can be :
cos(3x) = cos(2x+x)
= cos(2x)cos(x) - sin(2x)sin(x)
= (2cos^2 (x) - 1)cos(x) - 2sin(x)cos(x)sin(x)
= 2cos^3 (x) - cos(x) - 2(sin^2 (x))cos(x)

Answer 17

so far so good ?

Answer 18

sorry, there is typo for identity cos(2x), it should be
cos(2x) = 2cos^2 (x) - 1

Answer 19

yes

Answer 20

ok, then look the last part still there is sin^2 (x), we have to make all terms in cos, right ?

Answer 21

right

Answer 22

use the identity :
sin^2 (x) = 1 - cos^2 (x)

Answer 23

so, it can be
= ....
= 2cos^3 (x) - cos(x) - 2(sin^2 (x))cos(x)
= 2cos^3 (x) - cos(x) - 2(1 - cos^2 (x))cos(x)
= 2cos^3 (x) - cos(x) - 2cos(x) + 2 cos^3 (x)

Answer 24

just combine the similar terms :
2cos^3 (x) + 2cos^3 (x) = 4cos^3 (x)
- cos(x) - 2cos(x) = -3cos(x)

Answer 25

oh wow. thankyou

Answer 26

you're welcome :)