Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

How can I simplify sin^5x? I need to write it in terms of sinx, sin(3x) and sin(5x).

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
@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'm afraid you are :(
I am sorry but I actually have got to go bye

Not the answer you are looking for?

Search for more explanations.

Ask your own question