hartnn 3 years ago 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.

1. lgbasallote

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

2. hartnn

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

3. lgbasallote

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

4. lgbasallote

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

5. hartnn

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

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

7. joemath314159

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

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

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

10. ParthKohli

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

11. hartnn

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

12. joemath314159

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

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