Question

prove the identity

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

I think I understand a lot of it...but I need help though.

Answer 1

try using cos (a + b ) where a = x and b = 2x

Answer 2

so i should focus on the LHS? @chandanjha

Answer 3

just use the identity cos(a+b) = sina sinb+cosa cob

Answer 4

there should be a minus sign cross check it

Answer 5

start with the left and write it as

\[\cos(2x + x)\]

then use the expansion

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

next use the double angle information such as

sin(2x) = 2sin(x)cos(x)

and then its just processing to the answer on the right

Answer 6

in between sinb and cosa @chandanjha

Answer 7

use as @campbell_st has suggested

Answer 8

its cos a cos b - sin a sin b

Answer 9

ok

Answer 10

next use the double angle information such as

sin(2x) = 2sin(x)cos(x) i dont get this

Answer 11

@campbell_st

Answer 12

i wrote out the expansion but what do i dowith it?

Answer 13

this is a standard expansion of sin(2x)

sin(2x) = 2sin(x)cos(x)

so just substitute it and you get

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

which can be written as

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

next you need to look at the expansion of cos(2x)

there are several versions but a good starting point is

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

so now substitute this into

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

Answer 14

when you expand sin@x i dont see where the heck cos comes from

Answer 15

i dont see any identity...

Answer 16

its this

sin(2x) can be written as sin(x + x) using the expansion for the sum of 2 angles in sin, you get

sin(x +x) = sin(x)cos(x) + sin(x)cos(x) = 2sin(x)cos(x)


if you look at cos

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

Answer 17

ooooh so you have to like backtrack to go forward again?

Answer 18

not really

when studying this topic of trig equations, you start with sum and difference and then go to the double angle expansions....

some people memorise them, I just use the sum and difference but have the same angles...

its just stuff you need to know... to succeed in trig equations.

Answer 19

so I substituted it in, and got\[\cos ^{2}x-\sin^2xcosx-2\sin^2xcosx\] right?

Answer 20

Looks like you're in good hands.
Good luck.

Answer 21

not quite right

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

which becomes

\[\cos^3(x) - \sin^2(x)\cos(x) - 2\sin^2(x)\cos(x)\]

collect the like terms

and you'll get the right hand side...

Answer 22

why is cos to the third?

Answer 23

distribute

Answer 24

nvm

Answer 25

well you have to distribute cos(x)

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

which gives

\[\cos^3(x) - \sin^2(x)\cos(x)\]

Answer 26

thank youuuu!

Answer 27

These questions are just tedious.... just remember the sum and difference expansions and you can then work out the double angle information.

good luck

Answer 28

thanks