Question

show that sin 18
is the soluction to f(x)=8x^3-4x+1

Answer 1

without the use of a calfram

Answer 2

less of a calculus more of a trig ?

Answer 3

\[Hint :4\sin18*\cos36=1\]

Answer 4

\[x=18\]\[5x=90\]\[3x=90-2x\]\[\sin 3x=\cos 2x\]\[\sin x(3-4\sin^2 x)=\cos^2 x-\sin^2 x=1-2\sin^2 x\]\[y=\sin x\]\[y(3-4y^2)=1-2y^2\]
where did i do wrong?

Answer 5

\[4y^3-2y^2-3y+1=0\]

Answer 6

urs is correct too :)

Answer 7

anyway...i learned this method from @Callisto

Answer 8

jonask how u get that ???\[4 \sin 18 \cos 36=1\]

Answer 9

you used 3 instead of 8

Answer 10

\[4\cos36=\frac{ 1 }{ \sin18 }\]

Answer 11

it was an identity given and these is one of those hence question s do you think i should post the identity

Answer 12

easy solution will be veryfying by substitution
\[8\sin^318-4\sin18+1\]
\[4\sin18(2\sin^218-1)+1\]
\[4\sin18(-\cos36)+1\]
\[\frac{ 1 }{ \cos36 }(-\cos36)+1=-1+1=0\]

Answer 13

what method is sir @Callisto

Answer 14

what method did you show @mukushla

Answer 15

Referring to your question, actually, part b gives you a big hint.
\[1 = 4cos36sin18\]\[1= 4(1-2sin^218)sin18\]\[1= 4sin18-8sin^318\]Rearrange the terms and you'll get the equation in the form \[8x^2 - 4x + 1 =0\] I'm still thinking why the method introduced by @mukushla doesn't work.

Answer 16

wow thats relly an answer in disguise more than a Hint...well i see @mukushla printed
sinx(3-4sinx) wasn't it suppsedto be sinx(8-4sinx)

Answer 17

Hmm.. Usually for questions with part a, b, c and so on, the previous parts help you do the rest of the question.

As for mukushla's workings, he expanded sin3x to get that.
sin3x = sin(2x+x) = sin2x cosx + cos2x sinx
= (2sinx cosx) cosx + (1-2sin^2 x) sinx
= 2sinx cos^2x + (1-2sin^2 x) sinx
= 2sinx - 2sin^3 x + sinx - 2sin^3 x
= 3sinx - 4sin^3 x
= sinx (3-4sin^2 x)

Answer 18

i guess the method is fine as well but you get a different answer

Answer 19

answer that looks different but is the same i mean

Answer 20

That method would be fine, the only problem is how to get the equation in the form of 8x^3 - 4x +1 =0 :S

Answer 21

we need to show that 4y^3 + 2y^2 - y = 0
which is a problem by itself

Answer 22

Of course that's the problem, which we need to solve.

Answer 23

i meant that 4y^3 + 2y^2 - y this is what is different between the answers

Answer 24

I understand. But we have to show that 4y^3+2y^2-y =0.

Answer 25

the other problem is that sin 18 can be a solution of many other curves and it happens that @mukushla is one of them

Answer 26

@Jonask Actually, you just have to show that sin 18 is the solution to f(x)=8x^3-4x+1. It doesn't matter if it is a solution to other equation. And using the result of part b, you can do it. (well, I just show you).
We're thinking if there are no hints provided, which means we don't know the result of part b, and using the method introduced by mukushla here. How can we get the answer. Now, we get to the point which we have to prove 4y^3+2y^2-y = 0.

Answer 27

im giving up .. this method is wonderful anyway ive learned something

Answer 28

so we want to prove that when x=sin18

Answer 29

Yes. We set x=18 and y = sin18 in mukushla's work.

Answer 30

i notice that we changed \[4y^3-2y^2-3y+1=0\rightarrow 4y^3+2y^2-y=0\]

Answer 31

which also works for sin18

Answer 32

Not really.
Now, we have \(4y^3-2y^2-3y+1=0\) and we want to show that \(8y^3 - 4y +1 =0\)
So, add 4y^3 and subtract 2y^2 and y from the left of \(4y^3-2y^2-3y+1=0\), and at the same time, add 4y^3, 2y^2 and subtract y from the right of \(4y^3-2y^2-3y+1=0\). Then we get.

\(4y^3-2y^2-3y+1+4y^3+2y^2 -y= 4y^3 + 2y^2 -y\)
\(8y^3-4y+1= 4y^3 + 2y^2 -y\)

Compare \(8y^3-4y+1= 4y^3 + 2y^2 -y\) (something we have now) with \(8y^3 - 4y +1 =0\) (something we need)
We need to prove that 4y^3 + 2y^2 -y =0 (by looking at the right side of both equations), where y = sin18

Answer 33

@Jonask Btw, actually, you can solve you problem now. You don't have to solve it using this method.. :S

Answer 34

thanks,this is interesting thanks again

Answer 35

Welcome!~ This is very interesting indeed! :)