Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- hartnn

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- hartnn

- jamiebookeater

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- lgbasallote

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

- hartnn

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

- lgbasallote

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- lgbasallote

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

- 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.

- lgbasallote

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

- anonymous

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

- anonymous

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).

- hartnn

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

- ParthKohli

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

- hartnn

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

- anonymous

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

- mathslover

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.