## hartnn Group Title Aaah! new group for trigonometry :) Let me start this by asking to expand sin 5x in powers of sin x. Only for those who have started learning trigonometry and never tried this. Those who already know how to, can give other methods, not discussed here, later. one year ago one year ago

1. lgbasallote Group Title

using the new groups is fun, isn't it?

2. hartnn Group Title

its just the beginning...... Another similar exercise would be to write sin^5 x in terms of sin5x,sin4x,...

3. lgbasallote Group Title

...i never really learned these stuffs...all i know are the basics of trigonometry....

4. lgbasallote Group Title

but i do have an idea how to do it......but it's brute force.....is there a shorter way?

5. hartnn Group Title

there are few ways i am aware of...idk what do u mean by brute force here....u can try to expand here,if u want.

6. lgbasallote Group Title

my idea is using sin (2x) over and over

7. joemath314159 Group Title

hmm...the first way that pops into my mind is using:$e^{ix}=\cos(x)+i\sin(x)$and being a little clever.

8. joemath314159 Group Title

Noting that sin(5x)= Im(cos(5x)+isin(5x))=Im(e^(5ix))=Im((e^ix)^5) =Im((cos(x)+isin(x))^5). So basically, if you see what the imaginary part of:$(\cos(x)+i \sin(x))^5$is, you will have a formula for sin(5x).

9. hartnn Group Title

nice! lets see how new learners try it... maybe they don't even know that e^{ix} identity

10. ParthKohli Group Title

I do know that formula made by Euler, but I didn't know we could apply that here. :P

11. hartnn Group Title

u can try it parth.... maybe @mathslover also wanna try this....

12. joemath314159 Group Title

Here is an example of using Euler's formula to show that:$\cos(2x)=\cos^2(x)-\sin^2(x)$$\sin(2x)=2\sin(x)\cos(x)$

13. mathslover Group Title

Yeah , it is great to see the new pattern : @hartnn can you explain how to do it?