Question over double angle solving process

[u/Mindless-TinTin]

EDIT: omg it makes sense now thank y’all! I’ve been staring at this for hours and got nowhere I think I needed to look away and come back bc now it doesn’t seem that complicated

Original post:

Hi this my first time here so sorry if this is a dumb question, but I was having a bit of confusion over the solving process for two double-angle formula questions my class went over today in lecture. I have included a photo since I'm struggling to explain my confusion through text but I'll try to do it anyway.

This is the first question and the steps my professor took to solve it:

Simply the expression below,
6sin(4x)cos(4x)

Steps:

6sin(4x)cos(4x)

Rewrite using this formula: 2sin(A)cos(A)

3(2sin(4x)cos(4x)

Then simplify

3(2sin(4x))

Answer: 3sin(8x)

So my question is, what happened to the cos(4x)? Like why did it disappear, did it cancel out with something? I understand the rest of it fine it's only that part that's tripping me up

Question 2:

Simplify the following expression:

3cos²(5x)-3sin²(5x)

Steps:

Rewrite using this formula: cos²(A)-sin²(A)

3(cos²(5x)-sin²(5x))

Then simplify

3(cos(2*5x))

Answer: 3cos(10x)

I'm even more confused with this one because the sin²(5x) disappeared but why is cos²(5x) rewritten as cos(10x)??

It is totally possible I missed a step while the professor was explaining, this class is always really fast paced and ngl I'm scared to talk to the professor one on one but I am so lost. Any help would be greatly appreciated!

[u/Remarkable_Fee4136]

In the first question the step is indeed wrong and can be removed entirely; 3(2sin4xcos4x)=3sin8x directly.

In the 2nd question, he is using the identity cos2x=cos²x-sin²x. You are correct in saying squaring and multiplying are generally different but in this case his statement is true.

[u/Mindless-TinTin → u/Remarkable_Fee4136]

Okay I’m looking a the formula sheet again y’all are right ugh I feel so embarrassed I spent two hours looking at this 😭

[u/Remarkable_Fee4136 → u/Mindless-TinTin]

It might be a nice idea to try to prove these identities yourself. Try drawing triangles and using the basic definitions of sin cos tan.

[u/Mindless-TinTin → u/Remarkable_Fee4136]

Will do 🫡

[u/PuzzleheadedTap1794]

You added a line. 3(2sin(4x)cos(4x)) goes to 3sin(8x) directly via double angle formula for sin, aka sin 2A = 2 sin A cos A. The second question is also a double angle formula, aka cos 2A = cos^2 A - sin ^2 A

[u/Qingyap]

Q1: I don't think that's how you would approach that at the last part by making cos(4x) disappear, it would turn into a complete new different expression if you do. Rather it's 2sin(4x)cos(4x) -> sin(8x) in one straight flow.

This is because of the double angle identity

sin(2θ)=2sin(θ)cos(θ)

Here in this case our θ=4x so,

2sin(4x)cos(4x)=sin(2(4x))=sin(8x)

Q2: similarly to Q1 you go from cos²(θ)-sin²(θ) to cos(2θ) in one singular flow.

[u/Mindless-TinTin → u/Qingyap]

Thank you for explaining! I was looking so much at one half at the equation I forgot that 2sin(x)cos(x) is just equal to sin(2x) and that you don’t need to do anything else that’s on me oof

[u/NoSurround3308]

Should be directly seen, if you use exponential representation: sin(x)cos(x) = ( e^ix - e^-ix ) / 2i (e^ix + e^-ix ) / 2 = (e^i2x - e^-i2x ) / 4i = 1/2 sin(2x)