Use De Moivre's to express sin 4Theta

[charmedbeauty]

Homework Statement

Using De Moivre's formula to express sin 4θ in terms of sinθ and cosθ. Using this result, express sin4θcosθ in terms of sinθ only.

Homework Equations

The Attempt at a Solution

So sin4θ = [cosθ+isinθ]^4

using Binomial

...

cos4θ= cos^4(θ) -6cos^2(θ)(sin^2(θ)) +sin^4(θ)

&

sin4θ= 4cos^3(θ)(sinθ) -4cosθsin^3(θ)

so for the last part of the question I assume you replace the cos^2(θ) with 1-sin^2(θ)

and then sub this result into cos4θ from above.

But I am a little lost about "express sin4θcosθ in terms of sinθ only."

where does the sin4θcosθ come from?? Is this the right approach?? I have seen similar question but its just asked to represent sin4θ in terms of cosθ or visa-verca

the extra cos term is messing with my head

? Thanks.

 

[Curious3141]

Homework Statement

Using De Moivre's formula to express sin 4θ in terms of sinθ and cosθ. Using this result, express sin4θcosθ in terms of sinθ only.

Homework Equations

The Attempt at a Solution

So sin4θ = [cosθ+isinθ]^4

First of all, that should read cos4θ + isin4θ = [cosθ+isinθ]^4 .

Or alternatively, that sin4θ = Im([cosθ+isinθ]^4), where Im() signifies the imaginary part. (Right?)

using Binomial

...

cos4θ= cos^4(θ) -6cos^2(θ)(sin^2(θ)) +sin^4(θ)

&

sin4θ= 4cos^3(θ)(sinθ) -4cosθsin^3(θ)

OK so far.

so for the last part of the question I assume you replace the cos^2(θ) with 1-sin^2(θ)

and then sub this result into cos4θ from above.

Why bother? You're not asked to determine cos4θ. Just focus on sin4θ.

But I am a little lost about "express sin4θcosθ in terms of sinθ only."

where does the sin4θcosθ come from?? Is this the right approach?? I have seen similar question but its just asked to represent sin4θ in terms of cosθ or visa-verca

the extra cos term is messing with my head

? Thanks.

Let's use c to represent cosθ and s to represent sinθ.

You've established that sin4θ = 4sc^3 - 4cs^3.

Because you can only easily express *even* powers of c in terms of s using the identity c^2 = 1 - s^2, you'll have a problem with that expression, because of the odd powers of cosine.

That's why they wanted you to work on sin4θcosθ instead. When you mutliply it out, you get:

sin4θcosθ = 4sc^4 - 4c^2.s^3

so all your powers of cosine are even. Now regroup terms and apply that identity till you express everything in terms of s alone.

 

[charmedbeauty]

First of all, that should read cos4θ + isin4θ = [cosθ+isinθ]^4 .

Or alternatively, that sin4θ = Im([cosθ+isinθ]^4), where Im() signifies the imaginary part. (Right?)



OK so far.



Why bother? You're not asked to determine cos4θ. Just focus on sin4θ.



Let's use c to represent cosθ and s to represent sinθ.

You've established that sin4θ = 4sc^3 - 4cs^3.

Because you can only easily express *even* powers of c in terms of s using the identity c^2 = 1 - s^2, you'll have a problem with that expression, because of the odd powers of cosine.

That's why they wanted you to work on sin4θcosθ instead. When you mutliply it out, you get:

sin4θcosθ = 4sc^4 - 4c^2.s^3

so all your powers of cosine are even. Now regroup terms and apply that identity till you express everything in terms of s alone.

ohh it seems so obvious now!

I don't know if any other students have problems with this but one thing that seems to get me a lot of the time is reading the question and seeing what it is asking me to do.

Maybe I missed this part of math in high school, but I think that should be a focus in teaching since if I knew what the question was asking me I could have figured it out by myself.

Thanks for the help, greatly appreciated.