Find m , n , p constants

[CristianTheodor]

 

[MarkFL]

I think I would begin with the Pythagorean identity:

[MATH]\sin^2(x)=1-\cos^2(x)[/MATH]
Raise both sides to a power of 5, and arrange as:

[MATH]\sin^{10}(x)+\cos^{10}(x)=1-5\cos^2(x)+10\cos^4(x)-10\cos^6(x)+5\cos^8(x)[/MATH]
Now, use this form of the identity:

[MATH]\cos^2(x)=1-\sin^2(x)[/MATH]
Raise both sides to a power of 5, and arrange as:

[MATH]\sin^{10}(x)+\cos^{10}(x)=1-5\sin^2(x)+10\sin^4(x)-10\sin^6(x)+5\sin^8(x)[/MATH]
Adding our two results, we obtain:

[MATH]2(\sin^{10}(x)+\cos^{10}(x))=2-5(\sin^2(x)+\cos^2(x))+10(\sin^4(x)+\cos^4(x))-10(\sin^6(x)+\cos^6(x))+5(\sin^8(x)+\cos^8(x))[/MATH]
Arrange this as:

(1).....[MATH]10(\sin^4(x)+\cos^4(x))-10(\sin^6(x)+\cos^6(x))+5(\sin^8(x)+\cos^8(x))-2(\sin^{10}(x)+\cos^{10}(x))=3[/MATH]
Repeating this process, only raising to a power of 3, we obtain (after multiplying by -3):

(2).....[MATH]-9(\sin^4(x)+\cos^4(x))+6(\sin^6(x)+\cos^6(x))=-3[/MATH]
Adding (1) and (2), there results (after multiply by -7):

[MATH]-7(\sin^4(x)+\cos^4(x))+28(\sin^6(x)+\cos^6(x))-35(\sin^8(x)+\cos^8(x))+14(\sin^{10}(x)+\cos^{10}(x))=0[/MATH]
Add this to (1) to obtain:

[MATH]3(\sin^4(x)+\cos^4(x))+18(\sin^6(x)+\cos^6(x))-30(\sin^8(x)+\cos^8(x))+12(\sin^{10}(x)+\cos^{10}(x))=3[/MATH]
Divide by 3:

[MATH]\sin^4(x)+\cos^4(x)+6(\sin^6(x)+\cos^6(x))-10(\sin^8(x)+\cos^8(x))+4(\sin^{10}(x)+\cos^{10}(x))=1[/MATH]
And so, we find b) is the correct choice.

 

[Dr.Peterson]

That, of course, or something like it, is the right way.

Though I hate to say it, we could also use the fact that this is a multiple-choice question, and trust that one of the answers is correct, to find which it is.

Let x = 0 and find a simple relationship between m, n, and p; that will eliminate all but b and d.

You learn nothing more if x = pi/2

Then let x = pi/4, and you will find another relationship that eliminates d. So b has to be the answer, if anything is.

But that isn't math.

 

[CristianTheodor]

I think I would begin with the Pythagorean identity:

[MATH]\sin^2(x)=1-\cos^2(x)[/MATH]
Raise both sides to a power of 5, and arrange as:

[MATH]\sin^{10}(x)+\cos^{10}(x)=1-5\cos^2(x)+10\cos^4(x)-10\cos^6(x)+5\cos^8(x)[/MATH]
Now, use this form of the identity:

[MATH]\cos^2(x)=1-\sin^2(x)[/MATH]
Raise both sides to a power of 5, and arrange as:

[MATH]\sin^{10}(x)+\cos^{10}(x)=1-5\sin^2(x)+10\sin^4(x)-10\sin^6(x)+5\sin^8(x)[/MATH]
Adding our two results, we obtain:

[MATH]2(\sin^{10}(x)+\cos^{10}(x))=2-5(\sin^2(x)+\cos^2(x))+10(\sin^4(x)+\cos^4(x))-10(\sin^6(x)+\cos^6(x))+5(\sin^8(x)+\cos^8(x))[/MATH]
Arrange this as:

(1).....[MATH]10(\sin^4(x)+\cos^4(x))-10(\sin^6(x)+\cos^6(x))+5(\sin^8(x)+\cos^8(x))-2(\sin^{10}(x)+\cos^{10}(x))=3[/MATH]
Repeating this process, only raising to a power of 3, we obtain (after multiplying by -3):

(2).....[MATH]-9(\sin^4(x)+\cos^4(x))+6(\sin^6(x)+\cos^6(x))=-3[/MATH]
Adding (1) and (2), there results (after multiply by -7):

[MATH]-7(\sin^4(x)+\cos^4(x))+28(\sin^6(x)+\cos^6(x))-35(\sin^8(x)+\cos^8(x))+14(\sin^{10}(x)+\cos^{10}(x))=0[/MATH]
Add this to (1) to obtain:

[MATH]3(\sin^4(x)+\cos^4(x))+18(\sin^6(x)+\cos^6(x))-30(\sin^8(x)+\cos^8(x))+12(\sin^{10}(x)+\cos^{10}(x))=3[/MATH]
Divide by 3:

[MATH]\sin^4(x)+\cos^4(x)+6(\sin^6(x)+\cos^6(x))-10(\sin^8(x)+\cos^8(x))+4(\sin^{10}(x)+\cos^{10}(x))=1[/MATH]
And so, we find b) is the correct choice.

Thank you a lot