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BluFoot
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How can I simplify sin^5x? I need to write it in terms of sinx, sin(3x) and sin(5x).
 one year ago
 one year ago
BluFoot Group Title
How can I simplify sin^5x? I need to write it in terms of sinx, sin(3x) and sin(5x).
 one year ago
 one year ago

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BluFoot Group TitleBest ResponseYou've already chosen the best response.0
Hahaha I know how to use wolframalpha! I don't how to get there :P
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
Use Euler's formula. You know it?
 one year ago

BluFoot Group TitleBest ResponseYou've already chosen the best response.0
No, I'll look into it though and post back.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
\[\Im((e^{i\theta})^5)=\Im(e^{i5\theta})\]
 one year ago

BluFoot Group TitleBest ResponseYou've already chosen the best response.0
Definitely don't know that! This is a linear algebra course, the original question is find the coordinate vector for sin^5(x) from the basis {sinx,sin3x,sin5x}. So far, I've been trying to use addition rules to simplify them. sin(3x)=(sin2x+x) etc. But it's taking forever, there must be a better way!
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
\[e^{ix}=\cos(x)+\sqrt{1}\sin(x)\] \[(e^{ix})^2=(\cos(x)+\sqrt{1}\sin(x))^2=cos^2(x)sin^2(x)+2\sqrt{1}cos(x)sin(x)\] \[(e^{ix})^2=e^{i2x}=cos(2x)+\sqrt{1}sin(2x)\]Use the fact that if you have \[f(x)+\sqrt{1}g(x)=h(x)+\sqrt{1}p(x)\]Then this MUST be true: \[f(x)=h(x)\] and \[g(x)=p(x)\]
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
Use for\[e^{i5x}=(e^{ix})^5\]The \[\Im\]bit just means 'remove all the things that do not have a coefficient of \[\sqrt{1}\] in front of them.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt{1}=i=\text{ imaginary number } \] My method may look tricky, but it's MUCH simpler
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
http://betterexplained.com/articles/avisualintuitiveguidetoimaginarynumbers/
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
@henpen I think the equations you are using are beyond the level o math @blufoot is at I believe he is suppose to be using the base trig identites
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
*of
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
Fair enough, but it's always useful to know where the base identities come from. You help I need to sleep.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
Blufoot, you're on the right track, use sin(a+b)=sinacosb+sinbcosa
 one year ago

BluFoot Group TitleBest ResponseYou've already chosen the best response.0
Hey that's ok this is interesting! I remember learning a bit about imaginary numbers back in high school, but I don't see how it leads me to solving for sin(5x) :P I guess I'll just have to use the identities :( sin(3x) is easy, but sin(5x) takes forever because it's sin(3x+2x) which is cos(2x)sin(3x)+sin(2x)cos(3x)... So I need cos(3x) also and a ton of simplifying... I guess I'll give in to wolframalpha!
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
wait is this \[\sin ^{5}x\] or \[\sin(5x)\]
 one year ago

BluFoot Group TitleBest ResponseYou've already chosen the best response.0
Oh, either one, really. I'm trying to write sin^5x in terms of the other 3. I'm not sure if sin^5x is an easier one to simplify? Couldn't find a rule for that one...
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
there is a rule that may help with the load if you can figure out how to deal with the cos(x) its \[\sin ^{2}(x)=\frac{ 1\cos(2x) }{ 2 }\]
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
which I would think means \[\sin ^{5}(x)=\frac{ 1\cos(5x) }{ 5 }\] but I could also be horribly mistaken in that conclusion
 one year ago

BluFoot Group TitleBest ResponseYou've already chosen the best response.0
I'm afraid you are :(
 one year ago

whatisthequestion Group TitleBest ResponseYou've already chosen the best response.0
I am sorry but I actually have got to go bye
 one year ago
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