# Trigonometry Question

• Aug 9th 2009, 08:54 AM
Programmer
Trigonometry Question
If Sin[x] + Sin[x]^2 + Sin[x]^3 == 1
Find Cos[x]^6 + 4 Cos[x]^4 + 8 Cos[x]^2

Any help/solution would be appreciated.
Thanks.
• Aug 9th 2009, 02:56 PM
Is this the correct problem?

$\displaystyle \sin(x)+\sin^2(x)+\sin^3(x)=1$

$\displaystyle \cos^6(x)+4cos^4(x)+8cos^2(x)$
• Aug 9th 2009, 05:23 PM
Programmer
Quote:

Is this the correct problem?

$\displaystyle \sin(x)+\sin^2(x)+\sin^3(x)=1$

$\displaystyle \cos^6(x)+4cos^4(x)+8cos^2(x)$

I'm not too sure if the question is correct. I'll reconfirm it.
• Aug 9th 2009, 05:42 PM
Are you taking a highschool trig class? If so, lmk what topics you are covering. We're going to need that insight because this is a difficult problem and I'm not sure where to begin. Have you studied composite angle formulae? For example $\displaystyle \cos(2x)=1-2\sin^2(x)$
• Aug 10th 2009, 12:40 AM
Programmer
Quote:

Are you taking a highschool trig class? If so, lmk what topics you are covering. We're going to need that insight because this is a difficult problem and I'm not sure where to begin. Have you studied composite angle formulae? For example $\displaystyle \cos(2x)=1-2\sin^2(x)$

Yes, I have covered multiple angle formulas last year. A friend of mine gave this problem, I tried, but couldn't solve it.
• Aug 12th 2009, 07:15 AM
Quote:

Is this the correct problem?

$\displaystyle \sin(x)+\sin^2(x)+\sin^3(x)=1$

i think it has no solution . Neither sin (x) + sin (x^2) + sin (x^3) =1 has any solutions .
• Aug 12th 2009, 08:17 AM
alunw
I disagree. This is a continuous function which takes the value 0 at 0 and takes the value 3 at pi/2. So there is certainly some point between 0 and pi/2 where it is 1. It's somewhere between 30 and 36 degrees in fact.
Factorise the LHS as sin(x)(1+sin^2(x)) + sin^2(x))
Then you can get replace the sin^2 terms with 1-cos^2(x) and rearrange so that the remain sin(x) term is on its own. Then square and make the same subsitution. Now you have got rid of all the sin(x) terms and only have cos(x) terms left. Presumably after some messing about you can find the required function of cos(x).
• Aug 12th 2009, 08:21 AM
Programmer
Quote:

Originally Posted by alunw
I disagree. This is a continuous function which takes the value 0 at 0 and takes the value 3 at pi/2. So there is certainly some point between 0 and pi/2 where it is 1. It's somewhere between 30 and 36 degrees in fact.
Factorise the LHS as sin(x)(1+sin^2(x)) + sin^2(x))
Then you can get replace the sin^2 terms with 1-cos^2(x) and rearrange so that the remain sin(x) term is on its own. Then square and make the same subsitution. Now you have got rid of all the sin(x) terms and only have cos(x) terms left. Presumably after some messing about you can find the required function of cos(x).

I agree with the rest. Question is probably wrong, but that's what I was given, sorry about that.
- Wolfram|Alpha[x]+%2B+Sin[x]^2+%2B+Sin[x]^3+%3D%3D+1
shows 2 possible answers, both of which when plugged into the equation that was to be solved gave an irrational answer. The options for the answer were integers I'm sure (don't remember).
• Aug 12th 2009, 09:11 AM
alunw
I was merely pointing out that sin(x)+sin^2(x)+sin^3(x)=1 undoubtedly has a solution. The fact that is irrational is completely irrelevant. However the question probably is wrong, because if you follow the strategy I outlined this is what happens:
$\displaystyle sin(x)(1+sin^2(x)) + sin^2(x) = 1$
$\displaystyle sin(x)(1+1-cos^2(x)) + 1-cos^2(x) = 1$
$\displaystyle sin(x)(2-cos^2(x)) -cos^2(x) = 0$
$\displaystyle sin(x)(2-cos^2(x)) = cos^2(x)$
$\displaystyle sin^2(x)(4-4cos^2(x)+cos^4(x)) = cos^4(x)$
$\displaystyle (1-cos^2(x))(4-4cos^2(x)+cos^4(x)) = cos^4(x)$
$\displaystyle (4-4cos^2(x)+cos^4(x) -4cos^2(x)+4cos^4(x)-cos^6(x)) = cos^4(x)$
$\displaystyle (4-8cos^2(x)+4cos^4(x)-cos^6(x)) = 0$
$\displaystyle cos^6(x)-4cos^4(x)+8cos^2(x)=4$
• Aug 12th 2009, 09:49 AM
alunw
Further to my last post the angle, which is slightly under 33 degrees (you could find an exact expression by solving the cubic in cos^2(x) given by the final line above) is very close to, but not the same as, the angle at which $\displaystyle cos^6(x)+4cos^4(x)+8cos^2(x)=8$
and also to the angle at which $\displaystyle cos^4(x)=0.5$
but all three are different angles. There is less than 1/4 of a degree between the smallest and largest of these angles.