verifying identities: sin4x = (2sin^3(2x))(cos(2x)) + ....

[azdressagenut]

I am working on verifying trigonometric identities.

Directions: Verify that each trigonometric equation is an identity.

sin4x=(2sin[sup:31s6w5xu]3[/sup:31s6w5xu]2x)(cos2x)+(2sin2x)(cos[sup:31s6w5xu]3[/sup:31s6w5xu]2x)
=(2sin[sup:31s6w5xu]3[/sup:31s6w5xu]2x)(1-2sin[sup:31s6w5xu]2[/sup:31s6w5xu]x)+(2sin2x)(cos[sup:31s6w5xu]3[/sup:31s6w5xu]2x)

Im not sure if I have done the correct step here or not, also if I have I am not sure where to go from here.

tanxsin2x=2-2cos[sup:31s6w5xu]2[/sup:31s6w5xu]x

tanx(2sinxcosx)=2-2cos[sup:31s6w5xu]2[/sup:31s6w5xu]x

sinx(2sinxcosx)=2-2cos[sup:31s6w5xu]2[/sup:31s6w5xu]x
cosx

2cos[sup:31s6w5xu]2[/sup:31s6w5xu]x=2-2cos[sup:31s6w5xu]2[/sup:31s6w5xu]x

Is this verified at this point?

 

[Loren]

Re: verifying identities

I am working on verifying trigonometric identities.

Directions: Verify that each trigonometric equation is an identity.

sin4x=(2sin[sup:10wm6w9i]3[/sup:10wm6w9i]2x)(cos2x)+(2sin2x)(cos[sup:10wm6w9i]3[/sup:10wm6w9i]2x) <<< Did you start with sin4x = sin(2x+2x)?
=(2sin[sup:10wm6w9i]3[/sup:10wm6w9i]2x)(1-2sin[sup:10wm6w9i]2[/sup:10wm6w9i]x)+(2sin2x)(cos[sup:10wm6w9i]3[/sup:10wm6w9i]2x)

Im not sure if I have done the correct step here or not, also if I have I am not sure where to go from here.

tanxsin2x=2-2cos[sup:10wm6w9i]2[/sup:10wm6w9i]x

tanx(2sinxcosx)=2-2cos[sup:10wm6w9i]2[/sup:10wm6w9i]x

sinx(2sinxcosx)=2-2cos[sup:10wm6w9i]2[/sup:10wm6w9i]x <<< Looks to me like the cosx cancel leaving 2sin[sup:10wm6w9i]2[/sup:10wm6w9i]x. Now change to 1-cos[sup:10wm6w9i]2[/sup:10wm6w9i]x.
cosx

2cos[sup:10wm6w9i]2[/sup:10wm6w9i]x=2-2cos[sup:10wm6w9i]2[/sup:10wm6w9i]x <<< not a true statement.

Is this verified at this point?

 

[azdressagenut]

Re: verifying identities

No that is the original problem and I am supposed to work one side of the equation so it will match the other.

On the second problem if I cancel the cosx and change to 1-cos2x would that leave this:
2sin[sup:31rmasv3]2[/sup:31rmasv3]x = 1-cos2x ?

 

[soroban]

Re: verifying identities

Hello, azdressagenut!

We use this identity: .\(\displaystyle \sin2\theta \:=\:2\sin\theta\cos\theta\)

\(\displaystyle \sin(4x)\:=\:2\sin^3(2x)\cos(2x) + 2\sin(2x)\cos^3(2x)\)

\(\displaystyle \text{Factor the right side: }\;\underbrace{2\sin(2x)\cos(2x)}_{\text{This is }\sin(4x)}\,\underbrace{\left[\sin^2\!2x + \cos^2\!2x\right]}_{\text{This is 1}} \;=\;\sin(4x)\)

\(\displaystyle \tan x\sin2x \:=\:2-2\cos^2\!x\)

\(\displaystyle \text{The left side is: }\;\frac{\sin x}{\cos x}\cdot2\sin x\cos x \;=\;2\sin^2\!x \;=\;2(1-\cos^2\!x) \;=\;2-2\cos^2\!x\)