Here's the question you clicked on:
BluFoot
How can I simplify sin^5x? I need to write it in terms of sinx, sin(3x) and sin(5x).
Hahaha I know how to use wolframalpha! I don't how to get there :P
Use Euler's formula. You know it?
No, I'll look into it though and post back.
\[\Im((e^{i\theta})^5)=\Im(e^{i5\theta})\]
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!
\[e^{ix}=\cos(x)+\sqrt{-1}\sin(x)\] \[(e^{ix})^2=(\cos(x)+\sqrt{-1}\sin(x))^2=cos^2(x)-sin^2(x)+2\sqrt{-1}cos(x)sin(x)\] \[(e^{ix})^2=e^{i2x}=cos(2x)+\sqrt{-1}sin(2x)\]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 \[g(x)=p(x)\]
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.
\[\sqrt{-1}=i=\text{ imaginary number } \] My method may look tricky, but it's MUCH simpler
http://betterexplained.com/articles/a-visual-intuitive-guide-to-imaginary-numbers/
@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
Fair enough, but it's always useful to know where the base identities come from. You help- I need to sleep.
Blufoot, you're on the right track, use sin(a+b)=sinacosb+sinbcosa
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!
wait is this \[\sin ^{5}x\] or \[\sin(5x)\]
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...
there is a rule that may help with the load if you can figure out how to deal with the cos(x) its \[\sin ^{2}(x)=\frac{ 1-\cos(2x) }{ 2 }\]
which I would think means \[\sin ^{5}(x)=\frac{ 1-\cos(5x) }{ 5 }\] but I could also be horribly mistaken in that conclusion
I am sorry but I actually have got to go bye