[yellowsquirrel]

I am taking the FP2 exam this thurs and panicking! I need at least a B in f.maths and FP3 didn't go so well

I am trying to do some hyperbolic functions questions can anyone help?
1) given that sin(3x) = 3sin(x) - 4(sin^3(x)) and cos(3x) = 4(cos^3(x)) - 3cos(x) find expressions for sinh(3u) in terms of sinhu and cosh(3u) in terms of coshu

thank you

seriously stressing about these exams now

[Hasufel]

for the first identity, change the negative sign to a positive and change terms to hyperbolic, 2nd one, just change terms to hyperbolic (working on explaining why)

[f1mad]

Osborne's rule rule: product of two sine's, switch the sign for sinh.

[Emissionspectra]

Osborne's rule: YGM

[Hasufel]

(Original post by Emissionspectra)
Osborne's rule: YGM

big help!

here:

i`ll do the first one, even though your not meant to - then you might be able to do the 2nd:

we can write sinh(3x) as:

sinh(3x)= sinh(2x+x) = (sinh2x)(coshx)+(cosh2x)(sinhx)

=(2sinhx)(cosh^2(x))+sinh(x)+(co sh2x)(sinhx)

factorise here:

sinh(x)(2cosh^2(x)+cosh(2x))

= sinh(x)[2(1+sinh^2(x))+(1+2sinh^2(x)]

=sinh(x)(3+4sinh^2(x))
=
3sinh(x)+4sinh^3(x)

[Hasufel]

note: for hyperbolic functions, sinh(A+B) =sinhAcoshB+coshAsinhB

and cosh^2(x) = 1+2sinh^2(x)

these are the ones i`ve used for sinh(x) - use the corresponding ones for cosh

[aznkid66]

Of course, one can say that that the sinxsiny terms in trig identities become -sinhxsinhy terms in hyperbolic identities because of Osborne's rule, which you can directly apply for "simple" trig identity sin(3x) = 3sinx - 4sin^3(x) vs. sinh(3x) = 3sinhx + 4sinh^3(x).

[jack.hadamard]

(Original post by Hasufel)
note: for hyperbolic functions, sinh(A+B) =sinhAcoshB+coshAsinhB

and cosh^2(x) = 1+2sinh^2(x)

these are the ones i`ve used for sinh(x) - use the corresponding ones for cosh

You are supposed to use Osborne's rule.

Spoiler:

Show

Notice that you have



and




Hence, you find that and .

Now you see that , but that cosine remains the same.

Spoiler:

Show

Now, if you have



then, you could stick in to get



which simplifies to



the desired formula.

For the A-level exam, I suppose you could simply write down the answer and quote Osborne's rule.

[f1mad]

Sin(3x)= 3sinx - 4sin^3x

Notice: sin^3x= sinx*sinx*sinx

Since you have a product of two sin's there, you need to reverse it's sign to form an equation for sinh(3x).

Hence:

sinh(3x)= 3sinhx +4sinh^3(x)

[ztibor]

(Original post by yellowsquirrel)
I am taking the FP2 exam this thurs and panicking! I need at least a B in f.maths and FP3 didn't go so well

I am trying to do some hyperbolic functions questions can anyone help?
1) given that sin(3x) = 3sin(x) - 4(sin^3(x)) and cos(3x) = 4(cos^3(x)) - 3cos(x) find expressions for sinh(3u) in terms of sinhu and cosh(3u) in terms of coshu

thank you

seriously stressing about these exams now

Do you know the complex numbers, and the Euler's identity?
So for negative x

from these

similarly

Substituting ix=u and using that, by definition so f.e. you will get the formulas.

[Hasufel]

(Original post by jack.hadamard)
You are supposed to use Osborne's rule.

Spoiler:

Show

Notice that you have



and




Hence, you find that and .

Now you see that , but that cosine remains the same.

Spoiler:

Show

Now, if you have



then, you could stick in to get



which simplifies to



the desired formula.

For the A-level exam, I suppose you could simply write down the answer and quote Osborne's rule.

i have

http://mathworld.wolfram.com/OsbornesRule.html ALL OF THESE pertain to osborne the RULE - singular

question: find expressions for (NOT by direct substitution from osborne)
to OP: you choose what you think is easiest. That`s why people post hints here. (you`re not just meant to post formulas)

[aznkid66]

But obviously, if the question is only worth 1-2 marks, you need only apply Osborne's Rule.

[jack.hadamard]

(Original post by Hasufel)
question: find expressions for (NOT by direct substitution from osborne)

I don't like capital letters. I did not just post a formula, but showed to you how you derive it!

You seem to be missing the point of the question. It is all about Osborne's rule.
It asks you to provide the hyperbolic identity given the trigonometric identity.

[yellowsquirrel]

(Original post by Hasufel)
big help!

here:

i`ll do the first one, even though your not meant to - then you might be able to do the 2nd:

we can write sinh(3x) as:

sinh(3x)= sinh(2x+x) = (sinh2x)(coshx)+(cosh2x)(sinhx)

=(2sinhx)(cosh^2(x))+sinh(x)+(co sh2x)(sinhx)

factorise here:

sinh(x)(2cosh^2(x)+cosh(2x))

= sinh(x)[2(1+sinh^2(x))+(1+2sinh^2(x)]

=sinh(x)(3+4sinh^2(x))
=
3sinh(x)+4sinh^3(x)

Thanks so much I managed to do the other one now

Thanks everyone and good luck to the ppl taking it on thurs!